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Question

In an examination, there are 4 papers with a maximum of 'n' marks for each paper. Then the number of ways in which can one get 50% marks is?

A
13(n+1)(2n2+4n+3)
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B
13(n1)(2n2+4n+3)
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C
13(n+1)(2n24n+3)
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D
None of these
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Solution

The correct option is A 13(n+1)(2n2+4n+3)
Let the 4 papers be marked x,y,z,w
Let m be the maximum marks
0x,y,z,wm
x+y+z+w=2m
Number of solutions : 2m+41C41=2m+3C3
For, xm+1
Let t=x(m+1)t0
x+y+z+w=2mt+y+z+w=m1
Number of solutions: m1+41C41=m+2C3
For ym+1,zm+1,wm+1
Final solution : 2m+3C34×m+2C3
=13(m+1)(2m2+4m+3)
Hence the correct answer is 13(m+1)(2m2+4m+3)

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