JEE Main 2018 (Online) 15th April Evening ShiftQuestion 1
The coefficient of $x^{10}$ in the expansion of $(1+x)^{2}\left(1+x^{2}\right)^{3}\left(1+x^{3}\right)^{4}$ is equal to
Binomial Theorem · Coefficient of a Given Power / Term Independent of x
Solution
Concept. Take $x^a$, $x^{2b}$, $x^{3c}$ from the three factors and list all choices with $a+2b+3c=10$. Formulas used. (1) $(1+y)^n=\sum{}^{n}C_k\,y^k$Step 1 - Condition. $a+2b+3c=10$ with $0\le a\le2$, $0\le b\le3$, $0\le c\le4$; the term is ${}^{2}C_a\,{}^{3}C_b\,{}^{4}C_c$. Step 2 - List. $c=0$: $a+2b=10$ is impossible ($a+2b\le8$). $c=1$: $a+2b=7$: $(1,3)$ gives $2\cdot1\cdot4=8$. $c=2$: $a+2b=4$: $(0,2)$ gives $1\cdot3\cdot6=18$; $(2,1)$ gives $1\cdot3\cdot6=18$. $c=3$: $a+2b=1$: $(1,0)$ gives $2\cdot1\cdot4=8$. Step 3 - Add. $8+18+18+8=52$. Answer: (A)