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sunita123
I think i got the answer.
I thought this is asking for At least 5 members.


sunita123
A student committee that must consists of 5 members is to be formed from a pool of 8 candidates. How many different committees are possible
a 5
b 8
c 40
d 56
e 336

I am getting answer 93.
Pls help!!

They need to form committee of only 5 members, so out of 8 you need to pick only 5
YYYYYNNN, 5Y indicates selected candidates and 3N indicates those who were not selected...so that turns out to be simple arrangement problem with 5 like and 3 like terms - 8!/(5!*3!)
Hope it helps, regards
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sunita123
A student committee that must consists of 5 members is to be formed from a pool of 8 candidates. How many different committees are possible
a 5
b 8
c 40
d 56
e 336

I am getting answer 93.
Pls help!!


I am guessing you are getting 93 by doing this: 8C5 + 8C6 + 8C7 + 8C8 = 56 + 28 + 8 + 1 = 93
But note that the committee must consist of 5 members, not at least 5 members. Hence, all you have to do is 8C5 = 56

Answer (D)
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this is quite simple see as we have told a committee is to be selected it means the order of selection does not matter here

now we can select by this method,which calls slot method.

8x7x6x5x4/1x2x3x4x5= 56

hope its clear
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sunita123
A student committee that must consists of 5 members is to be formed from a pool of 8 candidates. How many different committees are possible

A. 5
B. 8
C. 40
D. 56
E. 336

Since we are selecting a committee, the order of selection does not matter and we have a combination.

We need to select 5 members from 8; thus 8C5:

8C5 = 8!/[5!(8-5)!] = 8!/(5!3!) = (8 x 7 x 6)/3! = (8 x 7 x 6)/(3 x 2) = 8 x 7 = 56

Answer: D
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Hi All,

This question can be solved with the Combination Formula:

N!/[K!(N-K)!] where N is the total number of items and K is the number in the sub-group.

We're told that we have to form a committee of 5 members from a total pool of 8 members, so N=8 and K=5. Using the Combination Formula, we have...

8!/[(5!(8-5)!] =

(8)(7)(6)(5)(4)(3)(2)(1) / (5)(4)(3)(2)(1)(3)(2)(1) =

(8)(7)(6) / (3)(2)(1) =

(8)(7) / 1 =

56 different combinations of 5 members.

Final Answer:

GMAT assassins aren't born, they're made,
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