How many ways to choose a committee of 3 people from 4 couples so that no couple is chosen.
A) 36
B) 48
C) 210
D) 306
E) None of these
GMAT Question of the Day: Permutation and Combination
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4 couples = 8 people
First spot can be filled in 8 ways
From remaining persons one is spouse so Second spot can be filled in 6 ways.
After second person selected remaining people are 6 out of which 2 are spouses of the first two so third spot can be filled in 4 ways.
Because order doesnt matter we need to divide by the number of combinations.
8X6X4/3x2X1 = 32 ways.
Answer is E — None of above.
I can be E only as the answer should be 32.
Choose a couple out of 4 to select 2 people (a couple) = 4C1 = 4
Choose the other one person out of the other 6 people 3 couples = 6C1 = 6
This makes a total of 6*4 = 24 ways in which a couple can be choosen.
Now, since there are 8 people altogether (4 couples), and we need to choose 3, there are a total of 8C3 combinations available = 56
So, the number of ways of choosing a team of 3 with no couple is
56-24 = 32 –> none of the above
ans is as follows.
when there is a negative condition we use the formula=total-positive condition.
total ways in which we can choose 3 people out of 8 is 8c3=56
positive= 1 couple+1 other person
=4c1*6c1=4*6=24
ans= 56-24=32
Another approach:
-Select three couples out of four that will be represented in a commitee. Total ways to do that: 4c1=4!/3!=4
-A represantative of a couple can be either one or another person (e.g. male or female). For each of three spots we have 2 ways to select a i.e. 2^3
4*2^3=32
E )
plz tell me the correct ans
E
Answer is E.
Get all combination for selecting 3 persons = 8c3 = 56
And the persons not included 1 couple + 1 person = 4c1 * 6c1 = 24
Now the team is 56 – 24 = 32
answer-E
Probability of selecting 1 person out of each couple
A1 A2 – com(2,1) => 2
B1 B2 – com(2,1) => 2
C1 C2 – com(2,1) => 2
D1 D2 – com(2,1) => 2
so (8) * prob of selecting 3 out of this 4 = 8*4 => 32
Ans is E.
However I am calculating some thing different:
for first choice we have 8 options.
second choice we have 6 options, as one person is already chosen and spouse of the chosen person cannot be selected.
third choice we have 4 options. same logic. 2 persons already chosen and 2 of their spouses cannot be selected.
8*6*4 = 192 ways
None of the above.
E None of these
Choose the couples to be represented first, then choose one member of each couple:
4C3 * 2C1 * 2C1 * 2C1 = 32
hi friends
lets chck no. of max to max way to select ppl
here we hv 4 men n 4 women
nw let comity contain only 4 men
so we can select a comity of 3 out 4 through =4c3=4
nw let comity contain only 4 women
so we can select a comity of 3 out 4 through =4c3=4
now
let comity contain 1 men n 2 woomen
so 1 men 4c1 n 2 women 3c2 (coz we dnt wana take 4 women coz it may be possible tht she should be wife of tht 1 man)
=12
n last
comity contain 2 man n 1 women
2 man by 4c2 way n 2 women by 2c2 way (same reason why we r taking only 2c2 ways)
=12
so tatal
=4+4+12+12
=32
=E
As order does not matter … 8*6*4/3!=32