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Q.
If the largest possible coefficient of $x^{2016}$ in $\left(1+x^{3 n}+x^{504}\right)^{10}$ (where $n \leq 25, n \in N$ ) is $c$, then the value of $n + c$ is
Binomial Theorem
Solution:
${ }^{10} C _0(1 \left.+ x ^{3 n}\right)^{10}+{ }^{10} C _1\left(1+ x ^{3 n}\right)^9 \cdot x ^{504}+{ }^{10} C _2\left(1+ x ^{3 n}\right)^8 \cdot x ^{1008}+{ }^{10} C _3\left(1+ x ^{3 n}\right)^7 x ^{1512}$
$ +{ }^{10} C _4\left(1+ x ^{3 n}\right)^6 x ^{2016}+\ldots \ldots$
Terms containing $x ^{2016}={ }^{10} C _3 \cdot{ }^7 C _{ r } x ^{3 nr +1512}+{ }^{10} C _4 \cdot{ }^6 C _0 \cdot x ^{2016}$
$3 nr =504 \Rightarrow nr =168 $
$n =\frac{168}{ r }, r =1,2,3, \ldots \ldots 6,7$
$n =24, r =7$
$\Rightarrow$ Coefficient $={ }^{10} C _3+{ }^{10} C _4={ }^{11} C _4$
$\Rightarrow n + c ={ }^{11} C _4+24={ }^{11} C _4+{ }^4 P _4 $