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296<br />

A Combinatorial Triad<br />

Cyrus Hsia<br />

student, University ofToronto<br />

One of the beautiful aspects of mathematics is that solving a problem<br />

can be done in so many ways. In combinatorics, the study of counting and<br />

ways of counting, this is more than evident in its many interesting problems.<br />

See any standard textbook on combinatorics [1] to convince yourself. Consider<br />

the following three standard combinatorial problems and see if you can<br />

nd the link between them.<br />

Problem 1. What is the number of ways to distribute 10 identical apples<br />

into 3 baskets labelled A, B, and C?<br />

Problem 2. How many ways are there to select 10 ice-cream scoops<br />

from 3 di erent avours onanice-cream cone?<br />

Problem 3. Find the number of non-negative integer solutions to the<br />

diophantine equation x1 + x2 + x3 =10.<br />

If you said the answersare all the same, then you arecorrect. The three<br />

problems above are di erent ways of counting something equivalent. Each<br />

way of thinking may be more helpful in certain situations than others. The<br />

following are solutions to approaching the apparently di erent, but similar,<br />

problems.<br />

Solution to Problem 1. The standard approach to solving this type of<br />

problem rests on the following key trick. Let the apples be represented by<br />

1's since they are all identical. Now use two 0's to place between the 1's<br />

to split them into 3 piles. Put the rst pile into basket A, the second into<br />

basket B and the remaining pile into basket C. The problem then reduces<br />

to nding the number of ways of placing the 0's among the 1's.<br />

There are10+1 = 11positions to place the rst 0, namely 9 spots<br />

between the 10 1's and the front and back of the list of 1's. Now there are<br />

11 numbers and 11+1 = 12positions to place the second 0. However,<br />

we are counting everything twice! If we place 0 in front and then a 0 after<br />

the fth 1, this is the same as if we place the rst 0 after the fth 1 and the<br />

second 0 in the rst spot. Hence, there are<br />

ways in total.<br />

11 12<br />

2<br />

Solution to Problem 2. This is exactly the same as Problem 1 if we<br />

consider the ice-cream scoops to be1's and the type of avour it is by its<br />

Copyright c 1998 <strong>Canadian</strong> <strong>Mathematical</strong> <strong>Society</strong>

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