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1. How many ways to choose a group of 3 people from 4 couples so that no couples are chosen?

Total combinations of choosing 3 out of 8 (order not important): 8!/(5!*3!) = 56
Combinations for groups with couples: 4 couples * 6 (one from each other couple) = 24
Total combinations - Combinations for groups with couples = 32

1. Three people can be chosen from a group of 8(4 couples) in 8C3 ways = 56 ways
Number of ways to choose 1 couple from a group of 4 is 4 * 6 - after choosing 1 couple there are 6 different ways to choose the other person.
So number of ways to choose a group of 3 people so that no couples are chosen will be 56 - 24 = 32 ways.

1. Three people can be chosen from a group of 8(4 couples) in 8C3 ways = 56 ways Number of ways to choose 1 couple from a group of 4 is 4 * 6 - after choosing 1 couple there are 6 different ways to choose the other person. So number of ways to choose a group of 3 people so that no couples are chosen will be 56 - 24 = 32 ways.

plz rthothad i don 't understand why you do retrieve 24 from 56; Plz can you explain why number of ways to choose a group of 3 people so that no couples are chosen will be 56 - 24 thanks

plz rthothad i don 't understand why you do retrieve 24 from 56; Plz can you explain why number of ways to choose a group of 3 people so that no couples are chosen will be 56 - 24 thanks

regards

mandy

mandy, My initial 8C3 combination contains couples also, so in order to remove all the combination that contains a couple, I had to deduct. Hope I am clear.