SOLUTION: out of 2n+1 students, n are to be awarded scholarship. number of ways in which at least one student gets scholarship is 63. find out how many students are recieving the scholarship

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Question 1043933: out of 2n+1 students, n are to be awarded scholarship. number of ways in which at least one student gets scholarship is 63. find out how many students are recieving the scholarship.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
I have been struggling to understand what the wording of the question means.
There are n%7D%7D+scholarships+available%2C%0D%0Aand+there+are+%7B%7B%7B2n%2B1 students applying for those scholarships.
Are those n scholarships distinguishable from each other?
Does "at least one student gets scholarship" mean n%3E=1 ?
Does the wording mean that maybe not all the n scholarships will be awarded, but at least 1 will be awarded?
The meaning of the wording has to allow the existence of 63 different possible outcomes. I could think of only one way to make that wording work.

Let's say the scholarships are indistinguishable from each other, and at least 1 , but at most n will be awarded:
If just 1 is awarded, that is combinations of 2n%2B1 taken 1 at a time,
and that is obviously 2n%2B1 ways,
because the single scholarship could be awarded to any of the 2n%2B1 students.
How many different ways could it happen that exactly 2 scholarship is awarded?
That is combinations of 2n%2B1 taken 2 at a time,
and that is %28matrix%282%2C1%2C2n%2B1%2Cn%29%29=%282n%2B1%292n%2F2=%282n%2B1%29n .
Similarly we would get to a number for the case that exactly 3 , 4 , %22...%22 , n scholarships could be awarded.
There are %28matrix%282%2C1%2C2n%2B1%2Cn%29%29=%282n%2B1%29%21%2F%28%28n%2B1%29%21n%21%29 possible different sets of n students that could be chosen to receive scholarships out of a pool of 2n%2B1 applicants.
Finding n so that the total of possible different sets of
1 , 2 , 3 , %22...%22 , or n scholarship recipients adds up to 63 appears complicated,
but the fact that 63 is a relatively small number makes it easy.
For n=4 , 2n%2B1=9 and there are
%28matrix%282%2C1%2C9%2C4%29%29=9%2A8%2A7%2A6%2F%282%2A3%2A4%29=126 possible different sets of 4 students that can be chosen from 9 applicants.
Since 126%3E63, we need to choose a smaller n .

It so happens that highlight%28n=3%29 works.
For n=3 , 2n%2B1=7 and from a pool of 7 students, there can be
%28matrix%282%2C1%2C7%2C3%29%29=7%2A6%2A5%2F%282%2A3%29=35 possible different sets of 3 students:
%28matrix%282%2C1%2C7%2C2%29%29=7%2A6%2F2=21 possible different sets of 3 students, and
7 sets of 1 student
that can be chosen to receive scholarships.
All in all there are 35%2B21%2B7=63 possible sets of 1 , 2, or 3 students that can be chosen to receive scholarships out of a total of 7 student applicants.