JEE Main 2024 — Binomial Theorem Question with Solution

$\begin{aligned} & x^2(1+x)^{98}+x^3\left(1+x^{97}\right)+x^4(1+x)^{96}+\ldots \ldots \\ & x^{54}(1+x)^{46} \end{aligned}$<br/>Coeff.of$\mathrm{x}^{70}:{ }^{98} \mathrm{C}_{68}+{ }^{97} \mathrm{C}_{67}+{ }^{96} \mathrm{C}_{66}+\ldots \ldots \ldots$ $\begin{aligned} & { }^{47} \mathrm{C}_{17}+{ }^{46} \mathrm{C}_{16} \\ & ={ }^{46} \mathrm{C}_{30}+{ }^{47} \mathrm{C}_{30}+\ldots \ldots \ldots . .{ }^{98} \mathrm{C}_{30} \\ & =\left({ }^{46} \mathrm{C}_{31}+{ }^{46} \mathrm{C}_{30}\right)+{ }^{47} \mathrm{C}_{30}+\ldots \ldots \ldots .{ }^{98} \mathrm{C}_{30}-{ }^{46} \mathrm{C}_{31} \\ & ={ }^{47} \mathrm{C}_{31}+{ }^{47} \mathrm{C}_{30}+\ldots \ldots \ldots . .{ }^{98} \mathrm{C}_{30}-{ }^{46} \mathrm{C}_{31} \\ & \ldots \ldots \\ & ={ }^{99} \mathrm{C}_{31}-{ }^{46} \mathrm{C}_{31}={ }^{99} \mathrm{C}_{\mathrm{p}}-{ }^{46} \mathrm{C}_{\mathrm{q}} \end{aligned}$<br/>Possiblevaluesof$(p+q)$are$62,83,99,46$ $\Rightarrow p+q=83$