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Question

The total number of non negative integral solution of the inequality 4i=1xi100

A
103C3
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B
104C4
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C
103C4
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D
104C3
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Solution

The correct option is D 104C4
Number of non negative solution of x1+x2+x3+x4100 will be same as the number of non negative solution of the equation
x1+x2+x3+x4+x5=100()
and the number of solutions of the equation is equal to the number of term in the expansion of
(x1+x2+...+x5)100 and the number of terms are 100+51C51=104C4
By using Number of terms in (x1+x2+...+xr)n=n+r1Cr1

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