How do you find the coefficient of #x^7" in "(1+x)^11#?

1 Answer
Ratnaker Mehta
Jul 4, 2016

Reqd. co-eff. #""11C_7=330#.

Explanation:

In the Expansion of #(a+b)^n#, the #(r+1)^(th)# term, denoted by #t_(r+1)#, is given by,

#t_(r+1)=""nC_r*a^(n-r)*b^r.#

Letting #a=1, b=x, n=11#, we have, #t_(r+1)=""11C_r*1^(11-r)*x^r="11C_r*x^r#.#

As we need the co-eff. of #x^7#, we have to take #r=7#, giving the reqd. co-eff. #""11C_7=11C_(11-7)=""11C_4={(11*10*9*8)/(1*2*3*4)}=330#.

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