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For a scholarship, at most n candidates out of 2n + 1 can be [#permalink]
17 Sep 2003, 01:52

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Question Stats:

0% (00:00) correct
0% (00:00) wrong based on 0 sessions

For a scholarship, at most n candidates out of 2n + 1 can be selected. If the number of different ways of selection of at least one candidate is 63, the maximum number of candidates that can be selected for the scholarship is:

Re: For a scholarship, at most n candidates out of 2n + 1 can be [#permalink]
26 Feb 2014, 09:18

Praetorian wrote:

For a scholarship, at most n candidates out of 2n + 1 can be selected. If the number of different ways of selection of at least one candidate is 63, the maximum number of candidates that can be selected for the scholarship is:

A. 3 B. 4 C. 2 D. 5

No of ways to select at least one candidate out of x candidates is 2^x - 1 2^x - 1 =63 2^x = 64 hence X= 6

But we are told that 2n+1 = x hence n = 2.5 or >2. Hence we choose A

_________________

“Confidence comes not from always being right but from not fearing to be wrong.”

Re: For a scholarship, at most n candidates out of 2n + 1 can be [#permalink]
26 Feb 2014, 18:58

Expert's post

ankur1901 wrote:

Praetorian wrote:

For a scholarship, at most n candidates out of 2n + 1 can be selected. If the number of different ways of selection of at least one candidate is 63, the maximum number of candidates that can be selected for the scholarship is:

A. 3 B. 4 C. 2 D. 5

No of ways to select at least one candidate out of x candidates is 2^x - 1 2^x - 1 =63 2^x = 64 hence X= 6

But we are told that 2n+1 = x hence n = 2.5 or >2. Hence we choose A

This is not correct. Getting a fractional value for n should be the clue since n = 2.5 could mean n = 2 candidates.

Remember that at most n candidates can be selected. The number of ways you can select at least 1 candidate up to n candidates out of the total 2n+1 is given as 63.

Note that you are given that (2n+1)C1 + (2n+1)C2 + ... + (2n+1)C(n) = 63. We also know (2n+1)C0 = 1. Also recall that (2n+1)C0 + (2n+1)C1 + (2n+1)C2 + ... + (2n+1)C(n) = (2n+1)C(n+1) + (2n+1)C(n+2) + ... + (2n+1)C(2n+1) since nCr = nC(n-r)

1 + 63 + 63 + 1 = 2^{2n+1} n = 3

For non-engineering students, this can be solved by backsolving though that's usually time consuming so I doubt GMAT will test such a concept.