09-23-2004, 05:37 PM
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#1 (permalink) |
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Within my grasp!
Join Date: Jun 2004
Location: New Delhi
Posts: 198
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Permutation combination ques.
A person wishes ot make up as many different parties as he can
out of his 20 friends such that each party consists of the same number of persons. How many friends should he invite???
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09-23-2004, 06:21 PM
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#2 (permalink) |
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TestMagic Guru-in-Training
 Join Date: Sep 2004
Posts: 606
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Re: Permutation combination ques.
If i have understan the question is it 10?
{ To maximize the fraction 20Cx= 20!/x!(20-x)!, where x is the number of people that he invites in each party. For x=1, 20C1=20. For x=2 20C2= (20)(19)/2. Notice that relative to 20C1 we have multiplied 20C1 by 19/2 > 1. Again 20C3 is bigger since we multiply 20C2 by the factor 18/3 > 1. To find the max of 20Cx w.r.t. x: this is 20C10 since 20C11=(20C10)(10/11) and 10/11<1 } |
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