Question about pairs of shoes

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?

 

Don't multiple post. :grumpy:

 

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

 

I don't know how to reach that answer either. I especially don't know how to reach an answer that has [itex]0 \choose 1[/itex] 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)/[itex]{10}\choose 4[/itex]
= 1 - [itex]5 \choose 4[/itex][itex]2 \choose 1[/itex]/210
= 1 - 1/21
= 20/21

 

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