A closet contains n pairs of shoes. If 2r shoes are chosen at random, (where 2r < n), what is the probability that there will be

a) no complete pair

b) Exactly one complete pair

c) Exactly 2 complete pair

We are choosing 2

2

r

shoes. How many ways are there to avoid a pair? The pairs represented in our sample can be chosen in (2)

(

n

2

r

)

ways. From each chosen pair, we can choose the left shoe or the right shoe. There are 22

2

2

r

ways to do this. So of the (22)

(

2

n

2

r

)

equally likely ways to choose 2

2

r

shoes, (2) 22

(

n

2

r

)

2

2

r

are "favourable."

Another way: A perhaps more natural way to attack the problem is to imagine choosing the shoes one at a time. The probability that the second shoe chosen does not match the first is 2-22-1

2

n

-

2

2

n

-

1

. Given that this has happened, the probability the next shoe does not match either of the first two is 2-42-2

2

n

-

4

2

n

-

2

. Given that there is no match so far, the probability the next shoe does not match any of the first three is 2-62-3

2

n

-

6

2

n

-

3

. Continue. We get a product, which looks a little nicer if we start it with the term 22

2

n

2

n

. So an answer is

22⋅2-22-1⋅2-42-2⋅2-62-3⋯2-4+22-2+1.

2

n

2

n

⋅

2

n

-

2

2

n

-

1

⋅

2

n

-

4

2

n

-

2

⋅

2

n

-

6

2

n

-

3

⋯

2

n

-

4

r

+

2

2

n

-

2

r

+

1

.

This can be expressed more compactly in various ways.