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Question

Find the number of terms in the expansion of (x1+x2+x3....xk)n


A

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B

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C

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Solution

The correct option is C


(x1+x2+x3......xk)n=r1+r2+r3...rk=nn!r1!+r2!×...rk!×xr11xr22....xrkk
Total numbers of terms in this expansion in this expansion is equal to the number of non-negative solution of r1+r2+...rk=n
=n+k1Ck1
=n+k1Cn


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