# Question about pairs of shoes

Alexsandro
I tryed to do this question of many ways, but I couldn't reach the answer. Could someone help me ?

"Four shoes are taken at random from five differents pairs. What is the probability that there is at least one pair among them" ?

The answer to this question is below, but I don't know how I reach it:

1- {$5 \choose 0$.$5 \choose 4$ + $5 \choose 1$.$4 \choose 3$ + $5 \choose 2$.$3 \choose 2$ + $5 \choose 3$.$2 \choose 1$ + $5 \choose 4$.$1 \choose 0$}/$10 \choose 4$.

Last edited:

## Answers and Replies

gjt01
if there are 10 shoes, and you take away 4, logically you would always have at least one pair left, but no more then three. Or am I understanding your question entirly wrong?

Staff Emeritus
Gold Member
Don't multiple post. :grumpy:

Homework Helper
gjt01 said:
if there are 10 shoes, and you take away 4, logically you would always have at least one pair left, but no more then three. Or am I understanding your question entirly wrong?

Yes, you are misunderstanding. He wants a pair in the shoes "taken", not in the shoes left.

Homework Helper
I don't know how to reach that answer either. I especially don't know how to reach an answer that has $0 \choose 1$ in it. The answer I got was:

P(at least 1 pair)
= 1 - P(no pairs)
= 1 - (# of ways to select 4 shoes with no pairs)/(# of ways to select 4 shoes)
= 1 - (# of ways to choose 4 pairs to select from * # of ways to pick one shoe from a given pair)/${10}\choose 4$
= 1 - $5 \choose 4$$2 \choose 1$/210
= 1 - 1/21
= 20/21

Alexsandro
AKG said:
I don't know how to reach that answer either. I especially don't know how to reach an answer that has $0 \choose 1$ in it. The answer I got was:

P(at least 1 pair)
= 1 - P(no pairs)
= 1 - (# of ways to select 4 shoes with no pairs)/(# of ways to select 4 shoes)
= 1 - (# of ways to choose 4 pairs to select from * # of ways to pick one shoe from a given pair)/${10}\choose 4$
= 1 - $5 \choose 4$$2 \choose 1$/210
= 1 - 1/21
= 20/21

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I wrote wrong, not is $0 \choose 1$, the correct is $1 \choose 0$. I repaired it.

Mithal
I found simpler way with the same answer as yours

From five pairs of shoes , choosing four pairs is 5 c 4

within the four pairs of shoes , we have 2 ^ 4 possibilities

Hence the number of combinations with no pairs chosen is =

(2^4) * 5 c 4 = 80

p ( at least one pair is chosen) = 1- (80)/ 10 c 4