The Handshake Problem RevisitedAuthor(s): David L. PagniSource: Mathematics in School, Vol. 20, No. 5 (Nov., 1991), pp. 43-44Published by: The Mathematical AssociationStable URL: http://www.jstor.org/stable/30216558 .

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Fig. 3

Sand

Marina

When all the groups were discussing their findings Marina sat quietly folding and cutting her square to form this upside down Napoleon's hat shape (Figure 3). When filled with sand this settles into a cone shape. Perhaps you'd like to try this on your own pupils. What would the calculated capacity be?

Geoff, Eddie and Brian

By simply folding the square to produce a container (Figure 4) like those you get in France for holding sweets (European dimension noted, I hope), this group obtained a capacity of approximately 250 ml. (Sand was used to measure this.) How would you calculate the capacity of this non-standard shape, given that the sand finds its own level at the top?

Fig. 4 Join edges

Sand

Notes

1. I have used capacity and volume as appropriate in this article. 2. SPREAD is one of the programs on the SMMIP (Secondary Math-

ematics with Micros Inservice Pack from Advisory Unit, Endymion Road, Hatfield).

3. I should be pleased to receive solutions to the above ideas for other possibilities, particularly from school pupils.

The handshake

problem

ted

by David L. Pagni, Professor of Mathematics, California State University

The Problem The "handshake problem" as it is called is becoming better and better known as a nice "hands on" activity that involves participants in problem solving. The problem goes like this:

If there are n people in a room and every person shakes hands with every other person in the room, how many total handshakes will there be in all?

By using a problem solving strategy called "examine a simpler case", students can form groups of 1, 2, 3, 4, 5, etc. and actually work through the problem by physically shaking hands and recording the results:

# of people in the group 1 2 3 4 5 6

# of handshakes 0 1 3 6 10 15

Mathematics in School, November 1991 43

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The process that seems most natural ior a particular group is to begin with a first person and have this person shake everyone's hand; a second person doesn't shake the first person's hand but does shake hands with the remaining people in the group; the third person does not shake hands with the first or second person but does with the rest, etc. ... until the next to last person shakes hands with just the last person.

For a group of 10 this results in the following number of handshakes:

9+8+7+6+5+4+3+2+ 1.

In general, for a group of n people this leads to the following sum:

(n- 1)+ (n - 2)+ (n - 3) + ... + 3 + 2+ 1. (1)

Analysis Students will sometimes use a different approach. For a group of ten people they think: "ten people, each person shakes hands with nine people, so there are 10 x 9= 90 handshakes; since there is duplication under this model which results in each pair of persons shaking hands twice, it is necessary to divide the answer by two.

10 x 9 The result is - = 45 handshakes".

2 For a group of n people this strategy leads to the formula

(n)(n- 1) 2(2)

as we think of n people shaking hands with n - 1 people and dividing by two to account for the duplication (x shakes hands with y and y shakes hands with x is really just one handshake).

The formula for the sum of the first k natural numbers, (kz)(k + 1)

reveals that formula (1) is equivalent to formula 2

(2). The problem can also be analysed using a geometric

model. If we think of n people about the room as points in a plane, then a handshake can be interpreted as drawing a segment to connect the points. (See Figure 1.)

The total number of segments determined by the ten points is thus the same as the number of handshakes. Since

two points determine a line (segment), this is the com- bination of n things (points) taken 2 at a time:

n! [(n)(n- 1)](n - 2)! (n)(n- 1) (n2) !(n- 2)! 2!(n- 2)! 2 2!(n - 2)! 2!(n - 2)! 2

There is also a set-subset interpretation. With n people in a room; a handshake is equivalent to forming a subset of size two (and having them shake hands). All possible subsets of size two that can be made for a set of size n is also

(n) (n - 1) (nz2 2

Extension A nice wrinkle to this problem is to ask, how many handshakes occur

(a) between two males; (b) between two females; (c) between a male and a female.

Suppose there are m males and females. Then, using the

(m)(m - 1) set-subset model, there are (m2) handshakes 2

(f)(f- 1) between males and (f2)= 2

handshakes between 2

females. Since there are m males and f females, each of m males

will shake hands with each of the females resulting in mxf handshakes between a male and a female. By acting out a simpler case between say, 2 boys and 3 girls, it becomes clear that there are 2 x 3 handshakes. Similarly, a tree diagram can help to visualise the procedure (Figure 2):

males

Fig.2

1

2

n

females 1 2 m

1 2

m

1 2 m

The result when all this is put together is a nice relation- ship between the total number of handshakes and the number of handshakes between males, between females, and between males and females:

Total # of # of handshakes # of handshakes # of handshakes handshakes = between males + between females + between males

and females

(m2 +f) = (m2) + (f2 + mf

This can be verified algebraically:

(m + f)! (m + f)(m +f- 1)

2!(m + f- 2)! 2

(m +f)2 _ -f M2 + 2mf + f2 - m -f 2 2

m(m - 1) f (f - 1) 2mf + + 2 2 2

= (m2) + (f2)+ mf

Mathematics in School, November 1991

Fig. 1 Six point diagram.

A

F E

D

Bt

Step 1 Draw all (5) unique segments "originating at point A".

Step 2 Draw all (4) unique segments "originating at point B".

Steps 3-6 Continue the process from the remaining points, the

C final step generating no new segments. [Note: The points are selected here at the vertices of a nonregular convex polygon. Does the placement of the points make a difference in modelling the problem? Explain.]

44

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