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Question

In an examination, the maximum marks for each of three papers are 50 each.

Maximum marks for the fourth paper are 100.

The number of ways in which the candidate can score 60% marks in the aggregate is


A

110556

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B

110500

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C

110356

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D

None of these

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Solution

The correct option is D

None of these


Explanation for the correct option:

Step-1: Solve for the required number of ways

The maximum marks that can be scored in all four papers=50+50+50+100=250

60% of the maximum aggregate is 60100×250=150 marks.

Let the marks scored in each paper be x1,x2,x3,x4 respectively

0≤x1,x2,x3≤50 and 0≤x4≤100

To score 150 marks in aggregate

x1+x2+x3+x4=150 subject to the above mentioned conditions

The number of ways of scoring 150

=coefficient of x150in1+x+x2+...+x5031+x+x2+...+x100 [marks scored can be any out of 0to 50 for first three and 0 to 100 for the last paper]

=coefficient of x150in 1-x511-x31-x1011-x ∵1+x+x2+...+xn=1-xn+11-x

=coefficient of x150 in 1-x5131-x1011-x-4

=coefficient of x150 in 1-3x51+3x102-x1531-x1011-x-4∵a-b3=a3-3a2b+3ab2-b3

= coefficient of x150 in (1−x153−3x51+3x102−x101−x254+3x152−3x203)1-x-4

Step-2: Solve further for the required number of ways

By using the binomial theorem we can write 1-x-n=1+nx+nn+12!x2+....

Substituting the value of n=-4 we get (1−x)-4=1+C14x+C25x2+....

The number of ways of scoring 150

=coefficient of x150in (1−x153−3x51+3x102−x101−x254−3x152−2x203)1+C14x+C25x2+....

Possible cases to get x150 are 1×x150 in second bracket, -3x51×x99 in second bracket, 3x102×x48 in second bracket, -x101×x49 in second bracket

The coefficient of xr in 1-x-nis Crn+r-1

Hence, the coefficient of x150=C150+4+150-1-3·C99+3·C48+4+48-14+99-1-1·C494+49-1

=C150+153-3·C99+3·C4851+102-1·C4952

=153!3!150!-3×102!99!3!+351!48!3!-52!49!3! ∵Crn=n!r!n-r!

=585276-3(171700)+3(20825)-22100

⇒ The coefficient of x150=110551

Thus the number of ways to score 60% marks aggregate are 110551.

Hence option(D) i.e. None of these is the correct answer.


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