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Less than half of a committee that is to be formed out of

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Less than half of a committee that is to be formed out of  [#permalink]

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New post 18 Mar 2013, 19:11
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Less than half of a committee that is to be formed out of certain group, are to be men. If there are 100 possible ways of forming the committee, then the number of men in the committee can be,

A. 2
B. 3
C. 4
D. 5
E. cannot be determined

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Re: Less than half of a committee that is to be formed out of  [#permalink]

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New post 18 Mar 2013, 20:13
SravnaTestPrep wrote:
Less than half of a committee that is to be formed out of certain group, are to be men. If there are 100 possible ways of forming the committee, then the number of men in the committee can be,

A. 2
B. 3
C. 4
D. 5
E. cannot be determined


100 possible ways means.
\(T^2\) = 100
so T= 10
So there are of total 10 members, making out of 100 members different ways.

Less than half of a committee that is to be formed out of certain group, are to be men.
m < \(1/2 T\)
m < 2.5
So m will be 2

Guys I have a doubt in this. If my abve method is correct, 100 possible ways means will give 10 by means of combinations or permutations ?
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Re: Less than half of a committee that is to be formed out of  [#permalink]

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New post 24 Mar 2013, 22:46
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1
SravnaTestPrep wrote:
Less than half of a committee that is to be formed out of certain group, are to be men. If there are 100 possible ways of forming the committee, then the number of men in the committee can be,

A. 2
B. 3
C. 4
D. 5
E. cannot be determined


1. Let there be m men and w women out of which the committee is formed.
2. mCn * wCr = 100, where n is the number of men chosen and r is the number of women chosen
3. The possible products for LHS of (2) are, 100 * 1, 50 * 2, 25 * 4, 20 * 5, 10 * 10
4. Since the number of women is greater than the number of men, the first three products in (3) can be rejected as they would not satisfy this condition.
5. In the case of the fourth product, 20 * 5 may be given by 6C3 * 5C1. Though this may satisfy the condition with 3 women and 1 man, 1 man is not in the choices given.
6. The last product 10 * 10 may be given by 5C2 * 5C3 with 2 men and 3 women. This satisfies the condition and also 2 men is one of the choices given.

Therefore the answer is choice A.
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Re: Less than half of a committee that is to be formed out of  [#permalink]

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New post 25 Mar 2013, 04:58
SravnaTestPrep wrote:
SravnaTestPrep wrote:
Less than half of a committee that is to be formed out of certain group, are to be men. If there are 100 possible ways of forming the committee, then the number of men in the committee can be,

A. 2
B. 3
C. 4
D. 5
E. cannot be determined


1. Let there be m men and w women out of which the committee is formed.
2. mCn * wCr = 100, where n is the number of men chosen and r is the number of women chosen
3. The possible products for LHS of (2) are, 100 * 1, 50 * 2, 25 * 4, 20 * 5, 10 * 10
4. Since the number of women is greater than the number of men, the first three products in (3) can be rejected as they would not satisfy this condition.
5. In the case of the fourth product, 20 * 5 may be given by 6C3 * 5C1. Though this may satisfy the condition with 3 women and 1 man, 1 man is not in the choices given.
6. The last product 10 * 10 may be given by 5C2 * 5C3 with 2 men and 3 women. This satisfies the condition and also 2 men is one of the choices given.

Therefore the answer is choice A.


"the first three products in (3) can be rejected as they would not satisfy this condition."
could not understand that y these 3 products r not satisfying the condition. :( as 100*1 can mean that there are 100 women and 1 man....50*2 means that 50 women and 2 men....25*4 means that 4 men 25 women.....

my other concern is how to calculate quickly that 20 is 6C3 or 10 is 5C2.

Please help me on this...Thanks
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Re: Less than half of a committee that is to be formed out of  [#permalink]

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New post 25 Mar 2013, 05:06
Perhaps wrote:
SravnaTestPrep wrote:
SravnaTestPrep wrote:
Less than half of a committee that is to be formed out of certain group, are to be men. If there are 100 possible ways of forming the committee, then the number of men in the committee can be,

A. 2
B. 3
C. 4
D. 5
E. cannot be determined


1. Let there be m men and w women out of which the committee is formed.
2. mCn * wCr = 100, where n is the number of men chosen and r is the number of women chosen
3. The possible products for LHS of (2) are, 100 * 1, 50 * 2, 25 * 4, 20 * 5, 10 * 10
4. Since the number of women is greater than the number of men, the first three products in (3) can be rejected as they would not satisfy this condition.
5. In the case of the fourth product, 20 * 5 may be given by 6C3 * 5C1. Though this may satisfy the condition with 3 women and 1 man, 1 man is not in the choices given.
6. The last product 10 * 10 may be given by 5C2 * 5C3 with 2 men and 3 women. This satisfies the condition and also 2 men is one of the choices given.

Therefore the answer is choice A.


"the first three products in (3) can be rejected as they would not satisfy this condition."
could not understand that y these 3 products r not satisfying the condition. :( as 100*1 can mean that there are 100 women and 1 man....50*2 means that 50 women and 2 men....25*4 means that 4 men 25 women.....

my other concern is how to calculate quickly that 20 is 6C3 or 10 is 5C2.

Please help me on this...Thanks



100 can mean only 100 C1, as you are selecting 1 woman out of 100 women. So you are selecting only 1 woman. 1 can mean only nCn and so you are selecting at least one man. But this does not satisfy the condition that the committee should have fewer men than women. Similarly other two cases can be ruled out.

I really do not know any short cut for figuring out how 20 is 6C3 and so on but you can try to remember them for smaller values.
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Re: Less than half of a committee that is to be formed out of  [#permalink]

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Re: Less than half of a committee that is to be formed out of   [#permalink] 08 Apr 2019, 21:26
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