permuations combinations

Reasoning and Quantitative Aptitude Permutation and Combination

1SITAMS, Chittoor

SREENIVASA INSTITUTE OF TECHNOLOGY ANDMANAGEMENT STUDIES

Murukambattu Post, Chittoor 517 127 (A.P)AFFILIATED TO JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY

ANANTAPUR,

Course materialFor

Reasoning and Quantitative Aptitude

Module Name: PERMUTATION AND COMBINATION


2SITAMS, Chittoor

PREQUISITIES:1. Factorial Notation:

Let n be a positive integer. Then, factorial n, denoted n! is defined as:

n! = n(n - 1)(n - 2) ... 3.2.1.

Examples:

i. We define 0! = 1.ii. 4! = (4 x 3 x 2 x 1) = 24.

iii. 5! = (5 x 4 x 3 x 2 x 1) = 120.2. Permutations:

The different arrangements of a given number of things by taking some or all at a time,are called permutations.

Examples:

i. All permutations (or arrangements) made with the letters a, b, c by taking two at atime are (ab, ba, ac, ca, bc, cb).

ii. All permutations made with the letters a, b, c taking all at a time are:( abc, acb, bac, bca, cab, cba)

3. Number of Permutations:

Number of all permutations of n things, taken r at a time, is given by:

nPr = n(n - 1)(n - 2) ... (n - r + 1) = )!(!rn

n

Examples:

i. 6P2 = (6 x 5) = 30.ii. 7P3 = (7 x 6 x 5) = 210.

iii. Cor. number of all permutations of n things, taken all at a time = n!.4. An Important Result:

If there are n subjects of which p1 are alike of one kind; p2 are alike of another kind; p3are alike of third kind and so on and pr are alike of rth kind,such that (p1 + p2 + ... pr) = n.

Then, number of permutations of these n objects is = )!)!....()(!(!

21 rpppn


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5. Combinations:

Each of the different groups or selections which can be formed by taking some or all of anumber of objects is called a combination.

Examples:

1. Suppose we want to select two out of three boys A, B, C. Then, possibleselections are AB, BC and CA.

Note: AB and BA represent the same selection.

2. All the combinations formed by a, b, c taking ab, bc, ca.3. The only combination that can be formed of three letters a, b, c taken all at a time

is abc.4. Various groups of 2 out of four persons A, B, C, D are:

AB, AC, AD, BC, BD, CD.

5. Note that ab ba are two different permutations but they represent the samecombination.

Number of Combinations:

The number of all combinations of n things, taken r at a time is:

nCr = )!)(!(!

rnr

n

= !)....2)(1(r

factprsrtonnn

Note:

.

nCn = 1 and nC0 = 1.i. nCr = nC(n - r)

Examples:

i) 11C4 = 330)1234(891011

ii) 16C13 = 16C(16 - 13) = 16C3 = 560123141516

!3141516


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SOLVED EXAMPLES:1. From a group of 7 men and 6 women, five persons are to be selected to form a committee so

that at least 3 men are there on the committee. In how many ways can it be done?

A)564 B)645 C) 735 D) 756 E) None of these

Answer: Option D

Explanation:

We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).

Required number of ways= (7C3 x 6C2) + (7C4 x 6C1) + (7C5)

=

1256

123567

16

12344567

+

1234534567

=(35x15)+(35x6)+21

=525+210+21

=756

2. When four fair dice are rolled simultaneously, in how many outcomes will at least one of thediceshow3?(1)155 (2)620 (3)671 (4) 625

Correct Answer - (3)

Solution:

When 4 dice are rolled simultaneously, there will be a total of 64 = 1296 outcomes.

The number of outcomes in which none of the 4 dice show 3 will be 54 = 625 outcomes.

Therefore, the number of outcomes in which at least one die will show 3 = 1296 625 = 671

3. In how many different ways can the letters of the word 'CORPORATION' be arranged so thatthe vowels always come together?


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A) 810 B) 1440 C) 2880 D) 50400 E)5760

Answer: Option D

Explanation:

In the word 'CORPORATION', we treat the vowels OOAIO as one letter.

Thus, we have CRPRTN (OOAIO).

This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.

Number of ways arranging these letters =!2!7

= 2520

Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged in!3!5= 20

ways.

Required number of ways = (2520 x 20) = 50400.

4. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can beformed?

A)210 B)1050 C) 25200 D) 21400 E) None of these

Answer: Option C

Explanation:

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)

= (7C3 x 4C2)

=

1234

123567

= 210

Number of groups, each having 3 consonants and 2 vowels = 210.

Each group contains 5 letters.


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Number of ways of arranging5 letters among themselves = 5!

= 5 x 4 x 3 x 2 x 1

= 120

Required number of ways = (210 x 120) = 25200

5. In how many ways can the letters of the word 'LEADER' be arranged?

A) 72 B) 144 C) 360 D) 720 E) None of these

Answer: Option C

Explanation:

The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.

Required number of ways = 360))(1!)(1!(2!)(1!6!

6. In a group of 6 boys and 4 girls, four children are to be selected. In how many different wayscan they be selected such that at least one boy should be there?


Answer: Option D

Explanation:

We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).

Required number of ways = (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)

= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2)

=

12564

123456

1234

1256)46(

= (24 + 90 + 80 + 15)


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= 209

7. How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which aredivisible by 5 and none of the digits is repeated?

A) 5 B) 10 C) 15 D) 20

Answer: Option D

Explanation:

Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1way of doing it.

The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5ways of filling the tens place.

The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways offilling it.

Required number of numbers = (1 x 5 x 4) = 20

8. In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 menand 10 women?


Answer: Option C

Explanation:

Required number of ways = (8C5 x 10C6)

= (8C3 x 10C4)

=

123478910

123678

=11760.


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9. A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls bedrawn from the box, if at least one black ball is to be included in the draw?


Answer: Option C

Explanation:

We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).

Required number of ways = (3C1 x 6C2) + (3C2 x 6C1) + (3C3)

= 161223

12563

= (45 + 18 + 1)

= 64

10. In how many different ways can the letters of the word 'DETAIL' be arranged in such a waythat the vowels occupy only the odd positions?

A) 32 B) 48 C) 36 D) 60 E) 120

Answer: Option C

Explanation:

There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.

Let us mark these positions as under:

(1) (2) (3) (4) (5) (6)

Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.

Number of ways of arranging the vowels = 3P3 = 3! = 6.

Also, the 3 consonants can be arranged at the remaining 3 positions.

Number of ways of these arrangements = 3P3 3! = 6.

Total number of ways = (6 x 6) = 36.


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11. In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3women?

A) 63 B) 90 C) 126 D) 45 E) 135

Answer: Option A

Explanation:

Required number of ways = (7C5 x 3C2) = (7C2 x 3C1) = 6331267

12. How many 4-letter words with or without meaning, can be formed out of the letters of theword, 'LOGARITHMS', if repetition of letters is not allowed?

A) 40 B) 400 C) 5040 D) 2520

Answer: Option C

Explanation:

'LOGARITHMS' contains 10 different letters.

Required number of words = Number of arrangements of 10 letters, taking 4 at a time.

= 10P4

= (10 x 9 x 8 x 7)

= 5040

13. In how many different ways can the letters of the word 'MATHEMATICS' be arranged sothat the vowels always come together?

A) 10080 B) 4989600 C) 120960 D) None of these

Answer: Option C

Explanation:

In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.


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Thus, we have MTHMTCS (AEAI).

Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest aredifferent.

Number of ways of arranging these letters = )!2)(!2(!8

= 10080

Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.

Number of ways of arranging these letters =!2!4

= 12.

Required number of words = (10080 x 12) = 120960

14. In how many different ways can the letters of the word 'OPTICAL' be arranged so that thevowels always come together?


Answer: Option B

Explanation:

The word 'OPTICAL' contains 7 different letters.

When the vowels OIA are always together, they can be supposed to form one letter.

Then, we have to arrange the letters PTCL (OIA).

Now, 5 letters can be arranged in 5! = 120 ways.

The vowels (OIA) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.

15.How many alphabets need to be there in a language if one were to make 1 million distinct 3digit initials using the alphabets of the language?

A) 26 B) 50 C) 100 D) 1000Correct Answer - (C)


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Solution:

1 million distinct 3 digit initials are needed.

Let the number of required alphabets in the language be n.

Therefore, using n alphabets we can form n * n * n = n3 distinct 3 digit initials.

Note distinct initials is different from initials where the digits are different.For instance, AAA and BBB are acceptable combinations in the case of distinct initials whilethey are not permitted when the digits of the initials need to be different.

This n3 different initials = 1 million

i.e. n3 = 106 (1 million = 106)=> n3 = (102)3 => n = 102 = 100

Hence, the language needs to have a minimum of 100 alphabets to achieve the objective.


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EXERCISE PROBLEMS:

1. In how many different ways may a class 1 spelunker who has never explored any of theeightcaves before set up a tour of three caves, if she wishes to explore caves Abbott andCaesar?

(a) 2 (b) 3 (c) 4 (d) 5 (e) 6

Ans. (b) Satyam Test Paper

2. Seven different toys are distributed among 3 children how many different ways arepossible?

(a) 7C3 (b) 7P3 (c) 3 7 (d) 7 3

Ans. (c) Wipro

3. Find the total number of distinct vehicle numbers that can be formed using twoletters followed by two numbers. Letters need to be distinct. TCS

Ans:65000

This question comes under permutations and combinations section. Out of 26 alphabetstwo distinct letters can be chosen in 26p2 ways. Coming to numbers part, there are 10 ways(any number from 0 to 9 can be chosen) to choose the first digit and similarly another 10ways to choose the second digit. Hence there are totally 10X10 = 100 ways. Combined withletters there are 6p2 X 100 ways = 65000 ways to choose vehicle numbers.

4. If the letters of the word SACHIN are arranged in all possible ways and thesewordsare written out as in dictionary, then the word SACHIN appears at serialnumber

( a ) 601 ( b ) 600 ( c ) 603 ( d ) 602 [ AIEEE 2005 ]

Ans: a)

if the word started with the letter A then the remaining 5 positions can be filled in `5! Ways

If it started with c then the remaining 5 positions can be filled in 5! Ways

Similarly if it started with H,I,N the remaining 5 positions can be filled in 5! Ways


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If it started with S then the remaining position can be filled with A,C,H,I,N inalphabetical order as on dictionary

The required word SACHIN can be obtained after the 5X5!=600 Ways

i.e. SACHIN is the 601th letter

5. How many ways are here to arrange the letters in the word GARDEN with thevowelsin alphabetical order ?

( a ) 120 ( b ) 240 ( c ) 360 (d ) 480 [ AIEEE 2004 ]

Ans : c

6. If repetition of the digits is allowed, then the number of even natural numbershavingthree digits is

( a ) 250 ( b ) 350 ( c ) 450 ( d ) 550 [ AIEEE 2002 ]

Ans: c)In a 3 digit number ones place can be filled in 5 different ways with (0,2,4,6,8)

10s place can be filled in 10 different ways

100s place can be filled in 9 different ways

There fore total number of ways = 5X10X9 = 450

7. If n+1C3 =2 nC2 , then the value of n is

( a ) 3 ( b ) 4 ( c ) 5 ( d ) 6 [ AIEEE 2002 ]

Ans: (c)

8. The number of arrangements of the letters of the word BANANA in which the twoNs do not appear adjacently is

( a ) 40 ( b ) 60 ( c ) 80 ( d ) 100 [ IIT 2002 ]


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Ans: (a)

9. Find the number of ways to arrange 4 people in groups of 3 at a time where ordermatters?

A)20 B)24 C)24 D)30

Answer:(B)

Explanation:

There are 24 ways to arrange 4 items taken 3at a time when order matters.

10. Find the number of ways to arrange 6 items in groups of 4 at a time where order matters?

A)36 B)720 C)360 D)420

Answer:(C)

Explanation:

There are 360 ways to arrange 6 items taken 4 at a time when order matters.

11. Find the number of ways to take 4 people and place them in groups of 3 at a time whereorder does not matter?

A)3 B)4 C)5 D)6

Answer:(B)

Explanation:

Since order does not matter, use the combination formula.


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Ans: (a)


A)20 B)24 C)24 D)30

Answer:(B)

Explanation:



A)36 B)720 C)360 D)420

Answer:(C)

Explanation:



A)3 B)4 C)5 D)6

Answer:(B)

Explanation:



14SITAMS, Chittoor

Ans: (a)


A)20 B)24 C)24 D)30

Answer:(B)

Explanation:



A)36 B)720 C)360 D)420

Answer:(C)

Explanation:



A)3 B)4 C)5 D)6

Answer:(B)

Explanation:



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There are 4 ways to arrange 4 items taken 3 at a time when order does not matter.

12. Find the number of ways to take 20 objects and arrange them in groups of 5 at a timewhere order does not matter.?

A)15504 B)40515 C)51540 D)15405

Answer : (A)

Explanation:

There are 15,504 ways to arrange 20 objects taken 5 at a time when order does not matter

13. How many ways are there to select a subcommittee of 7 members from among acommittee of 17?

A)19000 B)19500 C)19448 D)19844

Answer : (C)

Explanation:

Since it does not matter what order the committee members are chosen in, thecombination formula is used.

Committees are always a combination unless the problem states that someone like apresident has higher hierarchy over another person. If the committee is ordered, then itis a permutation.

C(17,7)= 19,448

14. Determine the total number of five-card hands that can be drawn from a deck of 52 cards.

A)2598960 B)2596800 C)2986800 D)2598708

Answer : ( A)


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A)15504 B)40515 C)51540 D)15405

Answer : (A)

Explanation:



A)19000 B)19500 C)19448 D)19844

Answer : (C)

Explanation:



C(17,7)= 19,448


A)2598960 B)2596800 C)2986800 D)2598708

Answer : ( A)


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A)15504 B)40515 C)51540 D)15405

Answer : (A)

Explanation:



A)19000 B)19500 C)19448 D)19844

Answer : (C)

Explanation:



C(17,7)= 19,448


A)2598960 B)2596800 C)2986800 D)2598708

Answer : ( A)


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Explanation:When a hand of cards is dealt, the order of the cards does not matter. If you are dealttwo kings, it does not matter if the two kings came with the first two cards or the last twocards. Thus cards are combinations. There are 52 cards in a deck and we want toknow how many different ways we can put them in groups of five at a time when orderdoes not matter. The combination formula is used.

C(52,5) = 2,598,960

15. Therefore there are 2,598,960 different ways to create a five-card hand from a deck of 52cards. A school has scheduled three volleyball games, two soccer games, and fourbasketball games. You have a ticket allowing you to attend three of the games. In howmany ways can you go to two basketball games and one of the other events?

A)6 B)5 C)30 D)11

Answer : (C)

Explanation :Since order does not matter it is a combination.

The word AND means multiply.

Given 4 basketball, 3 volleyball, 2 soccer.

We want 2 basketball games and 1 other event. There are 5 choices left.

C(n,r)

C(How many do you have, How many do you want)

C(have 4 basketball, want 2 basketball)*C(have 5 choices left, want 1)C(4,2)*C(5,1)(6)(5) = 30

Therefore there are 30 different ways in which you can go to two basketball games and oneof the other events.

16. How many ways are there to deal a five-card hand consisting of three eight's and twosevens?

A)5 B)24 C)42 D)54

Answer : (B)


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Explanation:If a card hand that consists of four Queens and an Ace is rearranged, nothing has changed.The hand still contains four Queens and an Ace. Thus, use the combination formula forproblems with cards.

We have 4 eights and 4 sevens.

We want 3 eights and 2 sevens.

C(have 4 eights, want 3 eights)*C(have 4 sevens, want 2 sevens)

C(4,3)*C(4,2) = 24

Therefore there are 24 different ways in which to deal the desired hand.

17. Find the number of ways to draw a straight, (suit does not matter) beginning with a 4 andending with a 8?

A)1024 B)1042 C)4012 D)8!Answer : (A)

Explanation:There are 5 slots.

__ __ __ __ __ The first slot must be a four. There are 4 ways to put a four in the first slot.There are 4 ways to put a five in the second slot, and there are 4 ways to put a six in the thirdslot. etc.

(4)(4)(4)(4)(4) = 45 = 1024

Therefore there are 1024 different ways to produce the desired hand of cards.

18. A certain marathon had 50 people running for first prize, second, and third prize.

A) How many different possible outcomes are there for the first three runners to cross thefinish line?

Solution:Order matters in a race, so use the permutation formula.

P(50,3) = 117,600 ways


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B) How many ways are there to correctly guess the first, second, and third place winners?

Solution:There is 1 way to correctly guess who comes in first, second, and third. There is only oneset of first, second and third place winners. You must correctly guess these three people,and there is only one way to do so.

19. A local delivery company has three packages to deliver to three different homes.if the packages are delivered at random to the three houses, how many ways are there

for at least one house to get the wrong package?A)3 B)5 C)3! D)5!

Answer : (B)

Solution:The possible outcomes that satisfy the condition of "at least one house gets the wrongpackage" are:One house gets the wrong package or two houses get the wrong package or three houses getthe wrong package.

We can calculate each of these cases and then add them together, or approach this problemfrom a different angle.

The only case which is left out of the condition is the case where no wrong packages aredelivered.

If we determine the total number of ways the three packages can be delivered and thensubtract the one case from it, the remainer will be the three cases above.

There is only one way for no wrong packages delivered to occur. This is the same aseveryone gets the right package. The first person must get the correct package and the secondperson must get the correct package and the third person must get the correct package.

1*1*1 = 1

Determine the total number of ways the three packages can be delivered.

3*2*1 = 6

The number of ways at least one house gets the wrong package is:

6 - 1 = 5

Therefore there are 5 ways for at least one house to get the wrong package.

20. Write down all the permutations of xyz.


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Sol:xyz, xzy, yxz, yzx, zxy, zyx.

21. How many permutations are there of the letters pqrs?

A)4 B)4^2 C)2^4 D)8Sol:4! = 1 2 3 4 = 24

22. .How many different four letter words can be formed (the words need not be meaningful)using the letters of the word "MEDITERRANEAN" such that the first letter is E and thelast letter is R?

A).59 B)!2!*2!*2

!11 C.56 D.23 E.!2!*2!*2!*3

!11

The correct choice is (A) and the correct answer is 59

Explanatory AnswerThe first letter is E and the last one is R.

Therefore, one has to find two more letters from the remaining 11 letters.

Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters.

The second and third positions can either have two different letters or have both theletters to be the same.

Case 1: When the two letters are different. One has to choose two different letters fromthe 8 available different choices. This can be done in 8 * 7 = 56 ways.

Case 2: When the two letters are same. There are 3 options - the three can be either Ns orEs or As. Therefore, 3 ways.

Total number of possibilities = 56 + 3 = 59

23. In how many ways can the letters of the word ABACUS be rearranged such that thevowels always appear together?

A.!2!6 B)3!*3! C)

!2!4 D.

!2!3!*4 E.

!2!3!*3

The correct choice is (D) and the correct answer is!2

!3!*4.

Explanatory Answer


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ABACUS is a 6 letter word with 3 of the letters being vowels.

If the 3 vowels have to appear together, then there will 3 other consonants and a set of 3vowels together.

These 4 elements can be rearranged in 4! Ways.

The 3 vowels can rearrange amongst themselves in!2!3

ways as "a" appears twice.

Hence, the total number of rearrangements in which the vowels appear together are!2

!3!*4

24. .In how many ways can the letters of the word "PROBLEM" be rearranged to make 7letter words such that none of the letters repeat?

A. 7! B)7C7 C)77 D)49The correct choice is (A) and the correct answer is 7!.

Explanatory AnswerThere are seven positions to be filled.

The first position can be filled using any of the 7 letters contained in PROBLEM.The second position can be filled by the remaining 6 letters as the letters should not repeat.The third position can be filled by the remaining 5 letters only and so on.

Therefore, the total number of ways of rearranging the 7 letter word = 7*6*5*4*3*2*1 = 7!Ways.

25. How many lines can you draw using 3 non collinear (not in a single line) points A, B andC on a plane?

A) 6 B) 4 C) 3 D)2

Solution:

You need two points to draw a line. The order is not important. Line AB is the same as lineBA. The problem is to select 2 points out of 3 to draw different lines. If we proceed as we didwith permutations, we get the following pairs of points to draw lines.

AB , AC

BA , BC

CA , CB


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There is a problem: line AB is the same as line BA, same for lines AC and CA and BC andCB.

The lines are: AB, BC and AC ; 3 lines only.

So in fact we can draw 3 lines and not 6 and that's because in this problem the order of thepoints A, B and C is not important.

26. How many ways can 10 letters be posted in 5 post boxes, if each of the post boxes cantake more than 10 letters?

(A)5^10 (B)10^5 (C)10P5 (D) 10C5Correct Answer - (A)

Solution:

Each of the 10 letters can be posted in any of the 5 boxes.

So, the first letter has 5 options, so does the second letter and so on and so forth for all of the10 letters.

i.e. 5*5*5*.*5 (upto 10 times)

= 5 ^ 10.

27. How many number of times will the digit 7' be written when listing the integers from 1to 1000?(A)271 (B)300 (C)252 (D) 304Correct Answer - (B)Solution:7 does not occur in 1000. So we have to count the number of times it appears between 1and 999. Any number between 1 and 999 can be expressed in the form of xyz where0 < x, y, z < 9.

1. The numbers in which 7 occurs only once. e.g 7, 17, 78, 217, 743 etcThis means that 7 is one of the digits and the remaining two digits will be any of the other

9 digits (i.e 0 to 9 with the exception of 7)

You have 1*9*9 = 81 such numbers. However, 7 could appear as the first or the secondor the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit, 2-digits and3- digits) in which 7 will appear only once.

In each of these numbers, 7 is written once. Therefore, 243 times.


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2. The numbers in which 7 will appear twice. e.g 772 or 377 or 747 or 77In these numbers, one of the digits is not 7 and it can be any of the 9 digits ( 0 to 9 withthe exception of 7).There will be 9 such numbers. However, this digit which is not 7 can appear in the first orm second or the third place. So there are 3 * 9 = 27 such numbers.

In each of these 27 numbers, the digit 7 is written twice. Therefore, 7 is written 54 times.

3. The number in which 7 appears thrice - 777 - 1 number. 7 is written thrice in it.

Therefore, the total number of times the digit 7 is written between 1 and 999 is243 + 54 + 3 = 300

28. . A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and theother only 4. In how many ways can they travel?(A)9 (B)26 (C)126 (D) 3920Correct Answer - (C)Solution:There are 8 students and the maximum capacity of the cars together is 9.

We may divide the 8 students as follows

Case I: 5 students in the first car and 3 in the second

Or Case II: 4 students in the first car and 4 in the second

Hence, in Case I: 8 students are divided into groups of 5 and 3 in 8C3 ways.

Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in 8C4 ways.

Therefore, the total number of ways in which 8 students can travel is8C3 + 8C4 = 56 + 70 = 126.

29. What is the value of 1*1! + 2*2! + 3*3! + ............ n*n!,where n! means n factorial or n(n-1)(n-2)...1(A)n(n-1)(n-1)! (B)(n+1)!/(n(n-1)) (C)(n+1)!-n! (D) (n + 1)! - 1!

Correct Answer - (D)Solution:

1*1! = (2 -1)*1! = 2*1! - 1*1! = 2! - 1!2*2! = (3 - 1)*2! = 3*2! - 2! = 3! - 2!3*3! = (4 - 1)*3! = 4*3! - 3! = 4! - 3!..

..


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..

n*n! = (n+1 - 1)*n! = (n+1)(n!) - n! = (n+1)! - n!

Summing up all these terms, we get (n+1)! - 1!

30. Eight first class and six second class petty officers are on the board of the 56 club. In

how many ways can the members elect, from the board, a president, a vice-president,

a secretary, and a treasurer if the president and secretary must be first class petty officers

and the vice-president and treasurer must be second class petty officers?

SOLUTION: Since two of the eight first class petty officers are to fill two differentoffices, we write

567.8!6!8

)!28(!88 2 P

Then, two of the six second class petty officers are to fill two different offices; thus, we write

305.6!4!6

)!26(!66 2 P

The principle of choice holds in this case; therefore, the members have56.30 = 1,680ways to select the required office holders

31. A standard deck of playing cards has 13 spades. How many ways can these 13 spades bearranged?A)13x13 B) 13 C) 13! D) (13+13)! E)None of these

Solution:

The solution to this problem involves calculating a factorial. Since we want to know how 13cards can be arranged, we need to compute the value for 13 factorial.13! = (1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13) = 6,227,020,800

32. A coach must choose five starters from a team of 12 players. How many different ways canthe coach choose the starters?A) 789 B) 987 C) 791 D) 792 E)297Answer:(D)Solution:


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Choose 5 starters from a team of 12 players. Order is not important.

33. There are fourteen juniors and twenty-three seniors in the Service Club. The club is to sendfour representatives to the State Conference. If the members of the club decide to send twojuniors and two seniors, how many different groupings are possible?A) 23000 B) 22032 C)23023 D) 22022 E) None of theseAnswer : (C)Solution:Choose 2 juniors and 2 seniors.

34.There are 2 brothers among a group of 20 persons. In how many ways can the group bearranged around a circle so that there is exactly one person between the two brothers?

(A)2*17! (B)18!*18 (C)19!*18 (D)2*18! (E)2*17!*17!

Correct Answer is 2 * 18! - Choice (D)ExplanationCircular Permutation 'n' objects can be arranged around a circle in (n - 1)!.

If arranging these 'n' objects clockwise or counter clockwise means one and the same, then thenumber arrangements will be half that number.

i.e., number of arrangements =2

)!1( n.

Let there be exactly one person between the two brothers as stated in the question.

If we consider the two brothers and the person in between the brothers as a block, then there will17 others and this block of three people to be arranged around a circle.

The number of ways of arranging 18 objects around a circle is in 17! ways.

Now the brothers can be arranged on either side of the person who is in between the brothers in2! ways.

The person who sits in between the two brothers could be any of the 18 in the group and can beselected in 18 ways.

Therefore, the total number of ways 18 * 17! * 2 = 2 * 18!.

35. How many integers, greater than 999 but not greater than 4000, can be formed with the digits0, 1, 2, 3 and 4, if repetition of digits is allowed?


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(A)499 (B)500 (C)375 (D)376 (E)501

Correct Answer is 376 - Choice (D)

Explanation

The smallest number in the series is 1000, a 4-digit number.

The largest number in the series is 4000, the only 4-digit number to start with 4.

The left most digit (thousands place) of each of the 4 digit numbers other than 4000 can take oneof the 3 values 1 or 2 or 3.

The next 3 digits (hundreds, tens and units place) can take any of the 5 values 0 or 1or 2 or 3 or 4.

Hence, there are 3 * 5 * 5 * 5 or 375 numbers from 1000 to 3999.Including 4000, there will be 376 such numbers.

36. In how many ways can 5 different toys be packed in 3 identical boxes such that no box isempty, if any of the boxes may hold all of the toys?

(A)20(B)30(C)25(D)600(E)480

Correct Answer is 25 - Choice (C)

Explanation:

The toys are different; The boxes are identical

If none of the boxes is to remain empty, then we can pack the toys in one of the following ways

a. 2, 2, 1b. 3, 1, 1

Case a. Number of ways of achieving the first option 2 - 2 - 1

Two toys out of the 5 can be selected in 5C2 ways. Another 2 out of the remaining 3 can beselected in 3C2 ways and the last toy can be selected in 1C1 way.

However, as the boxes are identical, the two different ways of selecting which box holds the firsttwo toys and which one holds the second set of two toys will look the same. Hence, we need todivide the result by 2.


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Therefore, total number of ways of achieving the 2 - 2 - 1 option is waysCC 152

3*102

23*25 Case b. Number of ways of achieving the second option 3 - 1 - 1

Three toys out of the 5 can be selected in 5C3 ways. As the boxes are identical, the remainingtwo toys can go into the two identical looking boxes in only one way.

Therefore, total number of ways of getting the 3 - 1 - 1 option is 5C3 = 10 = 10 ways.

Total ways in which the 5 toys can be packed in 3 identical boxes

= number of ways of achieving Case a + number of ways of achieving Case b= 15 + 10 = 25 ways.

37 . A key pad lock has 10 different digits, and a sequence of 5 different digits must be selectedfor the lock to open. How many key pad combinations are possible?A)24030 B) 30240 C) 30201 D) 12001 E)None of theseSolution:(B)

First we need to find n and r :

n is the number of digits we have to choose from. If n is 10 then there are 10 digits.

r is the number of digits being used at a time. If r is 5,then there are 5 digits in the sequence.Putting this into the permutation formula we get:

38 . How many necklaceof 12 beads each can be made from 18 beads of different colours?A) B) C) D) E)Ans. Here clock-wise and anti-clockwise arrangements are same.Hence total number of circularpermutations: 18P12/2x12 = 18!/(6 x 24)

39. In how many ways, can zero or more letters be selected form the letters AAAAA?A) B) C) D) E)Ans. Number of ways of :

A) Selectingzero'A's= 1


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3*102














26SITAMS, Chittoor


3*102














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Selecting one 'A's = 1Selecting two 'A's = 1Selecting three 'A's = 1Selecting four 'A's = 1Selecting five 'A's = 1

=> Required number ofways =6 [5+1]40. In how many ways can the letters of the word 'MISSISIPPI' be arranged?

a)1260b)12000c)12600d)14800 e) 26800Answer: (C )soution:

Total # of alphabets = 10

so ways to arrange them = 10!

Then there will be duplicates because 1st S is no different than 2nd S.

we have 4 Is 3 S and 2 Ps

Hence # of arrangements = 10!/4!*3!*2!

41. Goldenrod and No Hope are in a horse race with 6 contestants. How many differentarrangements of finishes are there if No Hope always finishes before Goldenrod and if all of

the horses finish the race?(A) 720(B) 360(C) 120(D) 24(E) 21Answer (B)Solution:two horses A and B, in a race of 6 horses... A has to finish before B

if A finishes 1... B could be in any of other 5 positions in 5 ways and other horses finishin 4! Ways, so total ways 5*4!

if A finishes 2... B could be in any of the last 4 positions in 4 ways. But the otherpositions could be filled in 4! ways, so the total ways 4*4!

if A finishes 3rd... B could be in any of last 3 positions in 3 ways, but the other positionscould be filled in 4! ways, so total ways 3*4!

if A finishes 4th... B could be in any of last 2 positions in 2 ways, but the other positionscould be filled in 4! ways, so total ways... 2 * 4!

if A finishes 5th .. B has to be 6th and the top 4 positions could be filled in 4! ways..


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A cannot finish 6th, since he has to be ahead of B

therefore total number of ways

5*4! + 4*4! + 3*4! + 2*4! + 4! = 120 + 96 + 72 + 48 + 24 = 360

42. If 5 nP3 = 4 n+1P3 find n?A)10 B) 12 C) 11 D)14Answer:(D)Solution:

nP3 = n(n 1) (n 2)n+1P3 = (n + 1) n (n 1) [Replacing n by (n + 1)] 5 n(n 1)(n 2) = 4(n +1)(n)(n 1)Divide both sides by n (n-1)5(n 2) = 4(n +1)5n 10 = 4n + 45n 4n = 4 +10

n = 1443. There are 7 non-collinear points. How many triangles can be drawn by joining these

points?

A) 35 B)10 C)8 D)7

Solution: A,A triangle is formed by joining any three non-collinear points in pairs.

There are 7 non-collinear pointsThe number of triangles formed = 7C3

=

!3

)27()17(7

=

123567

= 7 x 5= 35

44. How many diagonals can be drawn in a pentagon?B) 5 B)10 C)8 D)7

Answer:(A)

Solution. A pentagon has 5 sides. We obtain thediagonals by joining the vertices in pairs.Total number of sides and diagonals


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= 5C2

=

1245

= 5 x 2 = 10This includes its 5 sides also.Diagonals = 10 5 = 5Hence the number of diagonals = 10 5 = 5

45.From 8 gentlemen and 4 ladies, a committee of 6 is to be formed. In how manyways can this be done so that the committee contains exactly 2 ladies?A)6 B)70 C) 76 D)420Answer:(D)

Explanation: Exactly two ladies.

Ladies Gentlemen Number of ways4 82 4 4C2 x 8C4

But 4C2 = 6 and 8C4 = 70

The number of ways = 6 x 70 = 420

46) There are 6 bowlers and 9 batsmen in a cricket club. In how many ways cana team of 11 be selected so that the team contains at least 4 bowlers?

A)126 B)504 C)540 D)1170

Answer:(D)Explanation:

Possibilities Bowlers Batsmen Number of ways6 9

1 4 7 6C4 x 9C72 5 6 6C5 x 9C63 6 5 6C6 x 9C5

6C4 x 9C7 = 15 x 36 = 540

6C5 x 9C6 = 6 x 84 = 504

6C6 x 9C5 = 1 x 126 = 126

Total = 1170


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47. If 6Pr = 360 and 6Cr = 15 find r ?

A)4 B)3 C)6 D)5Answer:(A)

Explanation: nPr = nCr x r !6Pr = 15 x r!360 = 15 x r!

r! = 360/15= 24= 4 x 3 x 2 x 1

r! = 4!Therefore r = 4


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QUESTION BANK FOR MOCK TEST:

1. There are five women and six men in a group. From this group a committee of 4 is to bechosen. How many different ways can a committee be formed that contain three womenand one man?

A)55 B)60 C)25 D)192 [ ]

2. There are five women and six men in a group. From this group a committee of 4 is to bechosen. How many different ways can a committee be formed that contain at least threewomen?

A)65 B)67 C)20 D)12 [ ]

3. How many different 5-card hands include 4 aces?

A)61 B)28 C)48 D)225 [ ]

4. In a local election, there are seven people running for three positions. The person thathas the most votes will be elected to the highest paying position. The person with thesecond most votes will be elected to the second highest paying position, and likewise forthe third place winner. How many different outcomes can this election have?

A)301 B)258 C)176 D)210 [ ]

5. How many different ways can a five-question true-false test can be answered?

A)32 B)28 C)76 D)26 [ ]

6. How many ways can 10 people be placed in alphabetical order according to their firstnames?

A)1 B)10 C)100 D)10! [ ]

7. How many different ways can the letters in the word "STORE" be arranged?

A)15 B)5 C)120 D)24 [ ]

8. How many different ways are there to arrange the word SCHOOL?

A)455 B)53 C)487 D)360 [ ]

9. How many words with or without meaning, can be formed by using all letters of theword, DELHI, using each letter exactly once?


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A)10 B)25 C)60 D)120 [ ]

10. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can beformed?

A)210 B)1050 C)25200 D)21400 [ ]

11. In how many ways a committee, consisting of 5 men and 6 women can be formed from 8men and 10 women?

A)2660 B)5040 C)11760 D)86400 [ ]

12. In how many ways 7 different gems can be arranged in a necklace?

A)120 B)240 C)360 D)480 [ ]

13. A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 ballsbe drawn from the box, if atleast one black ball is to be included in the draw?

A)32 B)48 C)64 D)96 [ ]

14. How many 3-digit numbers are possible using the digits 0, 1, 4,5, 7 and 9?

A) 180 B) 120 C)100 D) 140 [ ]15. Find the number of combinations of 7 objects taking 2 at a time?

A) 12 B) 10 C) 21 D) 14 [ ]16. A class has 14 boys and 16 girls. The class has to elect one boy and one girl to be

representatives on a committee. How many different ways can they select them?

A) 114 B) 104 C) 224 D) 212 [ ]

17. How many arrangements of three types of flowers are there if there are 6 types to choosefrom?

A) 12 B) 18 C)20 D) 16 [ ]

18. If a university student has to choose 2 science classes from 5 available science classesand 3 other classes from a total of 7 other classes available, how many different groups ofclasses are there?

A) 225 B) 350 C) 175 D) 348 [ ]

19. How many ways can 7 people be arranged around a roundtable?


33SITAMS, Chittoor

A) 270 B) 720 C) 340 D) 625 [ ]20. A license plate has 3 letters and 3 digits in that order. A witnessto a hit and run accident

saw the first 2 letters and the last digit. If the letters and digits can be repeated, how manylicense platesmust be checked by the police to find the culprit?

A) 1300 B) 1800 C) 2600 D) 2200 [ ]

21. How many different ways can you have 1st, 2nd, 3rd, and 4thin a race with 10 runners?

A) 5040 B) 2800 C) 3684 D) 4840 [ ]

22. You select a president and vice-president from a group of 5students. Find the number ofpossible outcomes?

A) 18 B) 20 C) 28 D)24 [ ]

23. You must select a committee of 3 from 12 students. How manydifferent committees canbe formed?

A) 259 B) 360 C) 480 D) 720 [ ]

24. How many triangles can you make using 6 non collinear points on a plane?

A) 6C1=6 B) 6C4=15 C) 6C6=1 D) 6C3=20 [ ]

25. In how many ways can you arrange 5 different books on a shelf?

A) 5*5=25 B) 5!=120 C) 4!=24 D) 30 [ ]

26. Find the value of56

56

7777

PPPP

.

A) 1 B )2 C) 3 D) 6 [ ]

27. How many different ways can 4 red, 3 blue, 4 yellow, and 2 green bulbs be arranged ona string of Christmas tree lights with 13 sockets?

A) 900 B) 90 C) 90,90 D) 900,900 [ ]

28. On each trip, a salesman visits 4 of the 12 cities in his territory. In how many differentways can he schedule his route?

A)225 B)495 C)285 D) 197 [ ]


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29. How many permutations of 3 different digits are there, chosen from the ten digits 0 to 9inclusive?

A)84 B)120 C)504 D)720 [ ]30. How many permutations of 4 different letters are there, chosen from the twenty six letters

of the alphabet?

A) 14,950 B) 23,751 C)358,800 D) 456,976 [ ]31. A password consists of four different letters of the alphabet. How many different possible

passwords are there ?

A) 426 B)456,976 C)14,950 D)358,800 [ ]32. Restaurant offers 5 choices of appetizer, 10 choices of main meal and 4 choices of

dessert. A customer can choose to eat just one course, or two different courses, or allthree courses. Assuming all choices are available, how many different possible mealsdoes the restaurant offer?

A) 329 B)310 C)200 D)19 [ ]33. In a contest in which there are 8 participants, in how many ways can 5 distinct prizes be

Awarded?A.) 112 B.) 6720 C.) 336 D.) 672 [ ]

34. A club elects a president, vice-president, and secretary-treasurer. How many sets ofofficers are possible if there are 15 members and any member can be elected to eachposition? No person can hold more than one position.A.) 2730 B.) 32,760 C.) 910 D.) 1365 [ ]

35. A church has 7 bells in its bell tower. Before each church service 5 bells are rung insequence. No bell is rung more than once. How many possible sequences are there?

A.) 2520 B.) 42 C.) 84 D) 21 [ ]

36. How many arrangements can be made using 2 letters of the word HYPERBOLAS if noletter is to be used more than once?A.) 1,814,400 B.) 3,628,800 C.) 45 D.) 90 [ ]

37. A work softball team has 15 players on its roster. There are 9 distinct positions in whichthese players can be placed. How many lineups can be fielded?A.) 1,505,667,870 B.) 1,635,890 C.) 1,816,214,400 D.) 214,400 [ ]

38. From a group of 8 people, 5 will each win $1,000. How many different winning groupsare possible?

A.) 56 B.) 6720 C.) 168 D.) 336 [ ]


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39. Of a classroom filled with 20 students, 2 will be selected to stay after school and correcthomework for extra credit. How many combinations are possible?

A.) 190 B.) 210 C.) 63 D.) 40 [ ]

40. To win the lottery, one must correctly select 6 numbers from a collection of 50 numbers(one through 50). The order in which the selection is made does not matter. How manydifferent selections are possible?

A.) 250 B.) 15,890,700 C.) 300 D.) 13,983,816 [ ]

41. A test is administered with 15 questions. Students are allowed to answer any ten. Howmany choices of ten questions are there?

A.) 150 B.) 250 C.) 3003 D.) 3000 [ ]

42. There are 20 people who work in an office together. Four of these people are selected togo to the same conference together. How many such selections are possible?

A.) 116280 B.) 4845 C.) 3003 D.) 80 [ ]

43. There are 20 people who work in an office together. Four of these people are selected toattend four different conferences. The first person selected will go to a conference in NewYork, the second will go to Chicago, the third to San Franciso, and the fourth to Miami.How many such selections are possible?

A.) 116280 B.) 80 C.) 3003 D.) 4845 [ ]

44. Serial numbers for a product are to be made using three letters (using any letter of thealphabet) followed by two single-digit numbers. For example, JGR29 is one such serialnumber. How many such serial numbers are possible if neither letters nor numbers can berepeated

A.) 1,404,000 B.) 80 C.) 15690 D.) 117,000 [ ]

45. A 7-card hand is chosen from a standard 52-card deck. How many of these will have fourspades and three hearts

A.) 1001 B.) 204,490 C.)29,446,560 D.) 117,000 [ ]

46. In a new group of 15 employees at a restaurant, 10 are to be assigned as servers, 3 are tobe assigned as hosts, and 2 are to be assigned as cashiers. In how many ways can theassignment be made?


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A.) 60 B.) 204,490 C.)30,030 D.) 3014 [ ]

47. In how many ways can a first prize, a second prize and four identical third prizes beawarded to a group of 15 people?

A.) 60 B.) 150,150 C.)5005 D.) 3,603,600 [ ]

48. There are 10 students from whom 4 are going to be chosen to represent their school at aconference. If Jack, Anna or Chris, but only one of them, must be chosen, in how manyways can the students be chosen to go to the conference ?

A.) 210 B.) 150,150 C.)360 D.) 105 [ ]

49. There are 20 people who work in an office together. Four of these people are selected toattend four different conferences. The first person selected will go to a conference in NewYork, the second will go to Chicago, the third to San Franciso, and the fourth to Miami.How many such selections are possible ?

A.) 116,280 B.) 80 C.)4845 D.) 105 [ ]

50. If the letters of the word SACHIN are arranged in all possible ways and these wordsare written out as in dictionary, then the word SACHIN appears at serial number

( A ) 601 ( B ) 600 ( C ) 603 ( D ) 602 [ ]

ANSWERS

1 2 3 4 5 6 7 8 9 10

11

12

13

14

15

16

17

18

19

20

B A C D A A C D D C C C C A C C C B B C21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

A B D D B C D B D C D C B A A D C A A B41

42

43

44

45

46

47

48

49

50

C B A A B C B D A A

permuations combinations

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