JEE Main 2014 · 19 Apr · Shift 1 · Mathematics Q67

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MathematicsSingle correct+4 / -1Question 67 of 90

The coefficient of $x^{1012}$ in the expansion of $(1+x^n+x^{253})^{10}$ , (where $n\le22$ is any positive integer), is :

^{10}C_4

^{253}C_4

Answer and worked solution

Official answer: B

A term is $x^{nj+253k}$ with $j+k\le10$ . $1012=4\times253$ needs $k=4$ , $j=0$ . No other choice works: with $k\le3$ the rest, $nj=1012-253k\ge253$ , would exceed the largest possible $nj\le22\times10=220$ .

So the coefficient is $\dfrac{10!}{4!\,6!}={}^{10}C_4$ .

Topics:General and middle terms Combinations

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JEE Main 2014 · 19 Apr · Shift 1, Mathematics question 67