chapter 5 the binomial coefficients. summary pascal’s formula the binomial theorem identities...
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Chapter 5
The Binomial Coefficients
Summary
• Pascal’s formula
• The binomial theorem
• Identities
• Unimodality of binomial coefficients
• The multinomial theorem
• Newton’s binomial theorem
Review
• If k > n, C(n,k) = 0, C(n, 0) =1; If n is positive and 1≤k ≤n, then
• C(n,r) = C(n, n−r)
)!(!
!
!
)!/(!
),(
),(),(
rnr
n
r
rnn
rrP
rnP
r
nrnC
Pascal’s formula
• For all integers n and k with 1≤k ≤n-1,
• Hint: Let S be a set of n elements. We distinguish one of the elements of S and denote it by x. We then partition the set X of k-combinations of S into two parts , A and B such that all those k-combinations in A do not contain x while those in B contain x. Then
C(n, k) = |A| + |B| = C(n-1, k) + C(n-1, k-1).
1
11
k
n
k
n
k
n
Binomial Theorem
Let n be a positive integer. Then for all x and y,
In summation notation,
nnnnnn yyxn
nyx
nyx
nxyx
11221
121)(
n
k
kknn yxk
nyx
0
)(
Exercises
• Expand (x+y)5 and (x+y)6, using the binomial theorem.
• Expand (2x-y)7, using the binomial theorem.
Equivalent forms
n
k
knkn
n
k
knkn
n
k
kknn
yxk
nyx
yxkn
nyx
yxkn
nyx
0
0
0
)(
,)(
,)(
Special case
Let n be a positive number. Then for all x,
kn
k
kn
k
n xkn
nx
k
nx
00
1
Identities
• For positive integers n and k,
).0(,2
,22
21
1
,23120
,1
1
0
2
1
1
nn
n
k
n
nn
nn
nn
nnnn
k
nn
k
nk
n
k
n
n
Hints for proof
• Trivial ;• Set x=1 and y=-1 in the binomial theorem;• Differentiate both sides with respect to x for the
special case of binomial theorem and then substituting x=1;
• Counts the number of n-combinations of S (a set with 2n elements). Partition S into two subsets A and B. Each n-combination of S contains k elements of A and the remaining n-k elements in B. Note that C(n, k) = C(n, n-k).
Exercises
• Use the binomial theorem to prove that
• Generalize to find the sum for any real number r.
• Vandermonde convolution: for all positive integers mi, m2 and n,
.230
n
k
kn
k
n
n
k
krk
n
0
.0
2121
n
k n
mm
kn
m
k
m
Generalization
• Let r be any real number and k be any integer (positive, negative, or zero).
).1(
)0(
)1(
0
1!
)1()1(
kif
kif
kifk
krrr
k
r
Identities
For any real number r and integer k,
.1
1110
0
,1
1
1
0
k
n
k
n
k
n
kk
kfor
k
kr
k
krrr
Unimodality of binomial coefficients
Let n be a positive integer. The sequence of binomial coefficients is a unimodal sequence. More precisely, if n is even,
and if n is odd,
,1/210
n
n
n
n
n
nnn
.12/12/110
n
n
n
n
n
n
n
nnn
A corollary
• For n a positive integer, the largest of the binomial coefficients
.
n/2
n
n/2
n
n
n,,
2
n,
1
n,
0
n
is
Clutter
• Let S be a set of n elements. A Clutter of S is a collection C of combinations of S with the property that no combination in C is contained in another.
• Example: if S ={a, b, c, d} then
C = {{a, b}, {b, c, d}, {a, d} , {a, c}} is a clutter.
Sperner’s theorem
• Let S be a set of n elements. Then a clutter on S contains at most sets.
2/n
n
Multinomial coefficients
here n1,n2, …nt are non-nagative integers with n1+n2+ …+nt = n.
!!!
!
2121 tt nnn
n
nnn
n
Pascal’s formula for the multinomial coefficients
• Pascal’s formula for binomial coefficients:
• Pascal’s formula for multinomial coefficients
knk
n
knk
n
knk
n
k
n
k
n
k
n
1
1
1
1
1
11
.1
1
1
1
1
1
2121
2121
tt
tt
nnn
n
nnn
n
nnn
n
nnn
n
The multinomial theorem
Let n be a positive integer. For all x1, x2, …,xt,
where the summation extends over all non-negative integral solutions x1, x2, …,xt of x1+ x2+ …+xt = n.
tnt
nn
t
nt xxx
nnn
nxxx
21
2121
21 )(
Example and exercise
• When (x1+ x2+ …+x5)7 is expanded, the coefficient of x1
2x3x43x5 equals
• When (2x1 3x﹣ 2+5x3)6 is expanded, what the coefficient of x1
3x2x32 is?
.420!1!3!1!0!2
!7
13102
7
Newton’s Binomial Theorem
Let a be a real number . Then for all x and y with 0 ≤ |x| <|y|,
where
0
)(k
kkaa yxk
ayx
!
121
k
kaaaa
k
a
Special case
For any real number a,
k
k
a xk
ax
0
1
Correspondence
e.g. If a is a positive integer n, then when k>n0
k
n
kn
k
n xk
nx
0
1So
When a = -n
We can verify that
!k
knnn
k
n 11
!k
knnnk 111
1
11
11
n
kn
k
kn kk
For |y|<1
0 1
111
k
kkn yn
kny
Set xy
0 1
111
k
kkn xn
knx
0
11
1
k
kxn
kn
0 1
1
k
kxn
kn
For |y| < 1 and let n=1
0
1
0
1
1
11
k
k
k
kk
yy
yy
For a = 1/2
!k
k
k
1
21
321
221
121
21
21
!k
k
232
25
23
21
21
!k
kk
k
2325311 1
For a = 1/2 (cont’d)
!k
k
k k
k
2325311 1
21
!kk
kkk
k
2226422232543211 1
!
!
kk
kkk
k
213212221
1
1
1
22
21
12221
12
1
12
1
k
k
kkk
kk
k
k
k
!!
!
Special case for a =1/2
Consequently
k
kk
kk
k
xk
k
kx
kx
012
1
0
2
1
1
22
2
11 2
1
Assignments
• Exercises 6, 8, 11,15, 22, 34, 36 and 42 in the text book.
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