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A college admissions committee will grant a certain number of $10,000 scholarships, $5,000 scholarships, and $1,000 scholarships. If no student can receive more than one scholarship, how many different ways can the committee dole out the scholarships among the pool of 10 applicants? given that
problem number 1) there are two of each type
problem number 2) there are 1 x $10000, 2 x $50000, 3 x $1000
this question came to mind when i was solving a similar DS. what confused me was in my approach, the order of handing out scholarships seemed to matter. hopefully someone can clear my doubt
no OA. let the answers begin!!!
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Originally posted by amitdgr on 28 Aug 2008, 08:47.
Last edited by amitdgr on 28 Aug 2008, 21:13, edited 2 times in total.
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sset009
A college admissions committee will grant a certain number of $10,000 scholarships, $5,000 scholarships, and $1,000 scholarships. If no student can receive more than one scholarship, how many different ways can the committee dole out the scholarships among the pool of 10 applicants? given that
problem number 1) there are two of each type
problem number 2) there are 1 x $10000, 2 x $50000, 3 x $1000
this question came to mind when i was solving a similar DS. what confused me was in my approach, the order of handing out scholarships seemed to matter. hopefully someone can clear my doubt
A college admissions committee will grant a certain number of $10,000 scholarships, $5,000 scholarships, and $1,000 scholarships. If no student can receive more than one scholarship, how many different ways can the committee dole out the scholarships among the pool of 10 applicants? given that
problem number 1) there are two of each type
HHMMLL 1 2 3 4 5 6 7 8 9 10
=10C6 *6!/2!2!2! = 210* 90 = 18900 problem number 2) there are 1 x $10000, 2 x $50000, 3 x $1000
this question came to mind when i was solving a similar DS. what confused me was in my approach, the order of handing out scholarships seemed to matter. hopefully someone can clear my doubt
no OA. let the answers begin!!!
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x2suresh could you please explain the 6!/3!2 or 6!/2*2*2 bit? How do we get this? What is the logic behind this? Many thanks in advance!
x2suresh could you please explain the 6!/3!2 or 6!/2*2*2 bit? How do we get this? What is the logic behind this? Many thanks in advance!
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6! is the number of ways of arranging the 6 chosen people. But of the 6 when there are 3 of type1, 2 of type2 and 1 of type3.. then the actual number of arranging the 6 chosen people is 6! / 3!*2!*1!
when there are 2 of each type, then it becomes 6! / 2!*2!*2!
here the types represent the number of each type of schol available.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.