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a scholarship, at most

by sanju09 » Thu Jan 14, 2010 5:04 am
For a scholarship, at most n candidates out of 2n + I can be selected. If the number of different ways of selection of at least one candidate is 63, what is the maximum number of candidates that can be selected for the scholarship?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
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by ace_gre » Thu Jan 14, 2010 4:06 pm
I am struggling to get the concept behind this.. My crude way goes like this. :?

at most n candidates out of 2n+1

if n=2, 2n+1 = 5
if n=3, 2n+1 = 7
if n=4,2n+1=9....

atleast one candidate = no. of ways of selecting 1+ no. of ways of selecting 2 + no. of ways of selecting 3....

Starting with A, check if 5C1+5C2 equal to 63? No.
is 7C1+7C2+7C3 equal to 63? Yes. So n=3. So max number of candidates is (B)3

If it was not the case, proceed until n=6..

Is this correct? Please post the OA.

Thanks!
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by sanju09 » Fri Jan 15, 2010 12:46 am
good work ace_gre, you have already created and posted the OA.
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by imanemekouar » Fri Jan 15, 2010 6:59 am
Can you please explain the problem.I did not get the concept
Thank you so much
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by ace_gre » Fri Jan 15, 2010 1:41 pm
:D I am lucky to get this right..Not really sure how to explain it well..

For any 2n+1 candidates, n candidates can be selected atmost.
If atleast one candidate can be selected in 63 ways, then find maximum number of candidates n..

I started working from the options provided for n.

Created a table for n and 2n+1.

Now atleast one candidate can be selected => (2n+1)C1 + (2n+1)C2....(2n+1)Cn
Compute this for all values of 2n+1 and n provided.
For n=3, no. of ways of selecting atleast 1 candidate =63.

Hope this helps.
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