Bunuel wrote:
A group of 10 people consists of 3 married couples and 4 single men. A committee of 4 is to be formed from the 10 people. How many different committees can be formed if the committee can consist of at most 1 married couple?
(A) 105
(B) 207
(C) 210
(D) 540
(E) 5,040
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 fromSub-total : \(18+48+12=78\)
\(Total \ = \ 129+78 \ = \ 207\)
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