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06 Jun 2016, 04:53
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
51% (02:40) correct
49%
(02:51)
wrong
based on 531
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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
(A) 105
(B) 207
(C) 210
(D) 540
(E) 5,040
06 Jun 2016, 07:32
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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
= (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
06 Dec 2020, 02:17
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Bunuel
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 : \(18+48+12=78\)
\(Total \ = \ 129+78 \ = \ 207\)
