Q: There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?
Ans : 80
Ans : 80
Hey Jagdeep, first thing: please hide the answer option by using the spoiler tab. It allows the user to work on the question posted without any bias.
now coming to the question at hand:
#1 approach:
There are 3 spaces.
Space #1 can be any of the 10, because no one has been selected so there is no one to avoid.
Space #2 can choose 1 of only 8 people. 9 people left, but we cannot select the spouse of #1
Space #3 can choose 1 of only 6 people. 8 people left but we can't select the spouse of #1 or spouse of #2.
Now, we have 8 x 10 x 6 = 480 total possibilities.
This gives us the Permutation of the event, where we have identified order of #1, #2, #3, etc but the order doesn't matter. So we need to divide out the number of times (or permutations) of when we have the same people but in a different order. The total # of perms of a group of 3 is 3x2 = 6.
480/6 gives us the answer as 80.
Approach # 2
Total ways of selecting 3 people out of 10 = 10C3 = 10*9*8/3*2 = 120
In 3 people only 1 couple is possible. So way of selecting 1 couple out of 5 = 5C1 = 5
Way of selecting 1 person out of remaining 8 = 8C1 = 8
Total ways when there will be 1 couple and 1 different person = 8*5 = 40
Total number of ways in which no people will be from same couple = 120 - 40 = 80
Hope this helps!
