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Question

Coefficient of x25 in (1+x+x2+x3+...+x10)7 is:


A

31C16 - 7. 20C14

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B

31C14 - 7. 20C14

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C

31

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D

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Solution

The correct option is D


(1+x+x2+...+x10)7=(1x11)7(1x)7
=(17x11+21x2235x33+...)(1x)7
coefficient of x25=(coeff. of x257. coeff. of x14+21.coeff. of x3) in (1x)7
=31C257.20C14+21.9C3


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