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# For a scholarship, at most n candidates out of 2n + 1 can be

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CEO
Joined: 15 Aug 2003
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For a scholarship, at most n candidates out of 2n + 1 can be [#permalink]

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17 Sep 2003, 02:52
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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:

1. 3
2. 4
3. 2
4. 5
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18 Sep 2003, 07:11
Do u know the answer to the problem ?

I m not able to solve it, the qn is kinda confusing.
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18 Sep 2003, 17:36
I could do this only by backsolving.

63 = C(2n+1,1) + C(2n+1,2) + .... + C(2n+1,n)

Question stem is asking for n.

A. n = 3 then 2n+1 = 7.

7C1 + 7C2 + 7C3 = 7 + 21 + 35 = 63.

There has to be a better way, I think. Anyone? Vicks? Praet?
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19 Sep 2003, 02:56
edealfan wrote:
I could do this only by backsolving.

63 = C(2n+1,1) + C(2n+1,2) + .... + C(2n+1,n)

Question stem is asking for n.

A. n = 3 then 2n+1 = 7.

7C1 + 7C2 + 7C3 = 7 + 21 + 35 = 63.

There has to be a better way, I think. Anyone? Vicks? Praet?

Honestly, i cant think of a better way.
Vicks , can you help with this.
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Re: For a scholarship, at most n candidates out of 2n + 1 can be [#permalink]

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26 Feb 2014, 10:18
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
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Re: For a scholarship, at most n candidates out of 2n + 1 can be [#permalink]

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26 Feb 2014, 19:58
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.

$$(2n+1)C0 + (2n+1)C1 + (2n+1)C2 + ... + (2n+1)C(n) + (2n+1)C(n+1) + (2n+1)C(n+2) + ... + (2n+1)C(2n+1) = 2^{2n+1}$$

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.
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Re: For a scholarship, at most n candidates out of 2n + 1 can be   [#permalink] 26 Feb 2014, 19:58
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# For a scholarship, at most n candidates out of 2n + 1 can be

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