# 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:

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