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Q.
The coefficient of $x^{35}$ in the expansion of $\left(1+x^5\right)^{10}$ is
Binomial Theorem
Solution:
$ T _{ r +1}={ }^{10} C _{ r } \cdot x ^{5 r } ; r =0,1,2, \ldots \ldots ., 10$
$\therefore$ For coefficient of $x^{35}$, put $5 r=35$
$\therefore r =7$
So, $T _8={ }^{10} C _7 x ^{35} \Rightarrow$ Coefficient of $x ^{35}={ }^{10} C _7=\frac{10 !}{7 ! \cdot 3 !}=\frac{10 \times 9 \times 8}{6}=120$