# Question about pairs of shoes

1. Sep 20, 2005

### 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: Sep 21, 2005
2. Sep 20, 2005

### 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?

3. Sep 21, 2005

### Hurkyl

Staff Emeritus
Don't multiple post. :grumpy:

4. Sep 21, 2005

### HallsofIvy

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

5. Sep 21, 2005

### AKG

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

6. Sep 21, 2005

### Alexsandro

------------------------

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

7. Sep 21, 2005

### 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