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Difficulty:
35%
(medium)
Question Stats:
64% (01:13) correct
36%
(01:37)
wrong
based on 243
sessions
History
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Result
Not Attempted Yet
There are eight people in a certain club, including Bob and Ted. If the club is to select a 4 people committee that include Bob but not include Ted, how many different such committees are possible?
(A) 20
(B) 28
(C) 30
(D) 35
(E) 56
(A) 20
(B) 28
(C) 30
(D) 35
(E) 56
Just remove Bob and Ted from the calculation.
6C3 = 20
6C3 = 20
Using slot method:
We have to include Bob but exclude Ted, so we have to choose 4 people out of 8-1=7 (because Ted is no longer a candidate): Bobx6x5x4=120. Now, since order doesn't matter among the remaining 3 members, we have to exclude their arrangement possibilities as well by dividing 3!: 120/3!=120/6=20
Answer is A.
We have to include Bob but exclude Ted, so we have to choose 4 people out of 8-1=7 (because Ted is no longer a candidate): Bobx6x5x4=120. Now, since order doesn't matter among the remaining 3 members, we have to exclude their arrangement possibilities as well by dividing 3!: 120/3!=120/6=20
Answer is A.
Posts: 17
removing Bob and Ted we have to chose 3 people out of 6 and order does not matter
so 6C3 gives us 20 i.e. A
so 6C3 gives us 20 i.e. A
Posts: 158
mskx
but it says include Bob?
Posts: 4,849
sumankwan
Hi sumankwan.
From 8 people we have to choose 4.
Now Bob has to be there, so we have 7 people left out of which 3 have to be selected.
Ted is there in these 7 people, so he has to be excluded. Now we have 6 people left out of which 3 has to be chose = 6C3 = \(\frac{6 * 5 * 4}{3 * 2}\) = 20 possibilities
Option A
Hope this helps
Arun Kumar
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