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!

Thanks a lot buddy, very nice approach and i apologise for displaying the answer

fibbonnaci wrote: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!

jagdeep wrote: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

Please change the wordings if you are really looking for the various selections that ensure if a husband and wife may not be in the same group. I did it bold in your script for a reconsideration.

First draw can be made in 10 ways; second draw can be made in 8 ways, as we need to avoid the spouse of the person selected in the first draw. Third draw can be made in 6 ways now, as on this occasion we need to avoid 2 individuals, viz. the spouse of the person selected in the first draw and that in the second draw. In total, there could have been 10*8*6 ways for it, had it been the matter of arranging those three people on three distinct locations. But it is the case of mere selection, so let's covert an nPr to its nCr form by dividing the nPr result with r!, and (10*8*6)/3! = [spoiler]80[/spoiler].