Find the coefficient of $x^7$ in the expansion of $(1+x)^{11}$.

1. In $(1+x)^n$, the coefficient of $x^k$ is $\binom{n}{k}$ (since $T_{r+1} = \binom{n}{r}\,x^r$ — set $r = k$). 2. Here $n = 11$ and we want $x^7$, so $r = 7$. Coefficient = $\binom{11}{7}$. 3. Compute: $\binom{11}{7} = \dfrac{11!}{7!\,4!} = \dfrac{11\cdot 10\cdot 9\cdot 8}{4!} = \dfrac{7920}{24} = 330$. 4. Alternatively, using symmetry $\binom{11}{7} = \binom{11}{4} = 330$, so options B and C give the same numerical value but with different binomial coefficient notation. 5. Option A $\binom{11}{6} = 462$ is the coefficient of $x^6$ (off by one). Option D is off by another step. _Source: NCERT Class 11 Mathematics, Ch 7, §7.2.1 + Examples, p. 5–6._