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Permutation & Combination

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by shivanigs » Fri Jul 20, 2012 2:53 am
Hi,

Request your help to understand the logic behind this question.Thanks..


In how many ways can 3 people out of 14 people be given 3 prizes (gold,silver and bronze) so that Jack is awarded? Jack is one of the 14 people.

I don't have the OA for this one.
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by eagleeye » Fri Jul 20, 2012 3:25 am
shivanigs wrote:Hi,

Request your help to understand the logic behind this question.Thanks..


In how many ways can 3 people out of 14 people be given 3 prizes (gold,silver and bronze) so that Jack is awarded? Jack is one of the 14 people.

I don't have the OA for this one.
Hi shivanigs:

Since Jack must be awarded, he is always present among the prize winners. Then we need to select the other remaining 2 people from the 13 left for the other two prizes. No. of ways of selecting 2 people out of 13 = 13C2 = 13*6.
We are not done yet. We have 3 people but we can arrange the prizes between them in a number of ways.
No. of ways of awarding 3 people 3 prizes, when each gets only one = (choices for 1st prize)*(remaining choices for 2nd prize)*(last choice remaing after awarding 2 prizes) = 3*2*1=3!.

So total number of ways required= 13C2*3! = 13*6*6= 13*36= 468.
:)
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by karthik4vj » Fri Jul 20, 2012 11:33 pm
I used combination of both probablity and permutation. Not sure if it is right approach. Here we go ..

Step 1:
Number of ways to chose 3 winners out of 14 people = 14! / 11! = 2184

Step 2:
Probability that Jack is one among the winners = 3/14

Step 3:
Product of the above two steps should give u the answer = 2184 * (3/14) = 468

:)
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by shivanigs » Sat Jul 21, 2012 12:55 am
Thank you Eagle eye and Karthik for your responses,that really helped!
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