A group of 10 people consists of 3 married couples and 4 single men. A

Total number of possibilities of selecting 4 people out of 10 without any restrictions = 10C4 = 10!/(4!*6!)
= (10*9*8*7)/4*3*2*1
= 10*3*7
= 210
Number of possibilities of selecting 2 married couples = 3C2 = 3!/2! = 3
Number of different committees with atmost 1 married couple that can be formed = 210 - 3 = 207
Answer B
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The easier way -

Since 4 means 2 couples, let us find the ways in which we can select 2 couples = 3C2=3
Selecting 4 out of 10 without any restrictions = 10C4=\(\frac{10*9*8*7}{4*3*2}=210\)
Ways in which at most one couple is there = 210-3=207

Straight but lengthy method

I. No couple :
a) Remaining 4 from 4 individuals have to be chosen in 4C4 or 1 way
b) 3 from 4 individuals, and remaining 1 from the 3 couples = 3C1*2*4C3 = 3*2*4=24 ways...
3C1 because we can choose 1 couple from the 3 couples available, and 2 is because we have 2 people in that couple to choose from
c) 2 from 4 individuals, and remaining 2 from the 3 couples = 3C2*2*2*4C2 = 3*2*2*6=72 ways...
3C2 because we can choose 2 couple from the 3 couples available, and 2 is because we have 2 people in that couple to choose from
d) 1 from 4 individuals, and remaining 3 from the 3 couples = 3C3*2*2*2*4C1 = 2*2*2*4=32 ways...
3C3 because we can choose 3 couple from the 3 couples available, and 2*2*2 is because we have 2 people in each of the three couples to choose from
Sub-total : \(1+72+24+32=129\)

II. One couple :
a) Remaining 2 from 4 individuals = 3C1*4C2 = 3*6=18 ways...
3C1 because we can choose 1 couple from the 3 couples available.
b) 1 from 4 individuals, and remaining 1 from the other two couples = 3C1*4*4C1 = 3*4*4=48 ways...
3C1 because we can choose 1 couple from the 3 couples available, and 4 is because we have 2 people in each couple to choose from, that is 2*2 or 4 to choose from
c) Remaining 2 from the other two couples = 3C1*2*2 = 3*2*2=12 ways...
3C1 because we can choose 1 couple from the 3 couples available, and 2 is because we have 2 people in first couple to choose from, and 2 is because we have 2 people in second couple to choose from
Sub-total : \(18+48+12=78\)

\(Total \ = \ 129+78 \ = \ 207\)
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Shouldn't the possibilities of selecting 2 married couples be 4C2 here? Please explain.

There are only 3 married couples. Number of ways 2 couples can be selected out of 3 is 3C2.

If there are no restrictions on how to select the 4 people from a group of 10 people, then we have:

10C4 = 10!/[4!(10-4)!] = 10!/(4!6!) = (10 x 9 x 8 x 7)/4! = (10 x 9 x 8 x 7)/(4 x 3 x 2 x 1) = 5 x 3 x 2 x 7 = 210

ways to select them.

All of these ways will consist of at most 1 married couple, except if the 4 people picked consist of 2 married couples. So, let’s determine the number of ways 2 married couples can be picked as a committee of 4:

If a committee of 4 consists of 2 married couples, then it could be: (couple 1, couple 2), (couple 1, couple 3), or (couple 2, couple 3). Thus, there are only 3 ways 2 married couples can be picked for the committee of 4. Subtract this from 210 and we have 207 ways to select a committee of 4 that will consist of at most 1 married couple.

Answer: B
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Why are we subtracting the number of married couple from total? Why not add them to 210? Would be happy to know the reasoning behind this:)

We are subtracting the number of married couple from the total because we want to form a committee that has at most 1 married couple.
The total number of ways by which we can form a committee of 4 from 10 people(4 single and 3 couples) is 210. These 210 ways can have different combinations of single men and couples.

But the number of couples should not exceed 1, that is it should not be 2. So in how many ways it can exceed 1 = 3C2. Meaning from 3 couples we are selecting 2 couples for the committee such that the committee will have all couples and no single men. We want this case excluded. Therefore 210 - 3C2 = 210 - 3 = 207.

Hi nick1816 Bunuel ,

committee with atmost 1 married couple = committee with no married couple+committee with 1 married couple
so,
4C4(single men alone, non married couple)+3C1*4C2
1+18
19

please tell me whats wrong in this approach.

I do have the same question.
I think In the highlighted part 3C1 and 4C2 don't signify what you are trying to convey.
3C1 means selecting 1 out of 3
4C2 means selecting 2 out of 4

chetan2u can you help?!!
Doesn't 'at most' means we can have no couple also?
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Here, the highlighted part is, selecting 1 couple from 3 couple, which gives us 2 persons (3C1)
then selecting 2 persons from 4 single men (4C2)
so totally 4 persons we have to select.

Yes, it does mean so. But none of the married couple means there can be one of the two, but not both.
So if A1-A2, B-1-B2, C1-C2 are the three couples, and four others are w, x, y and z. Then no couples would also include A1-B1-C1-x or A2-B1-C1-x, and just not w-x-y-z
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