Properties of Binomial Coefficients — JEE Main Maths PYQ MCQs with Solutions

Solution: By the binomial theorem, $(1+1)^n = \sum_{r=0}^{n} \binom{n}{r} = 2^n$.

Solution: By the binomial theorem, $(1-1)^n = \sum_{r=0}^{n} (-1)^r \binom{n}{r} = 0^n = 0$ for $n \geq 1$.

Solution: The property $\binom{n}{r} = \binom{n}{n-r}$ implies that if $\binom{n}{r} = \binom{n}{s}$, then either $r = s$ or $r = n - s$ (i.e., $r + s = n$).

Solution: From $(1+1)^n = 2^n$ and $(1-1)^n = 0$, adding: $2(\binom{n}{0} + \binom{n}{2} + \binom{n}{4} + \ldots) = 2^n$, so even sum = $2^{n-1}$. Then odd sum = $2^n - 2^{n-1} = 2^{n-1}$.

Solution: The identity $\sum_{r=0}^{n} \binom{n}{r}^2 = \binom{2n}{n}$ follows from $(1+x)^n (1+x)^n = (1+x)^{2n}$, comparing coefficients of $x^n$.

Solution: Using $\sum_{r=0}^{n} (r+1) \binom{n}{r} = \sum_{r=0}^{n} r \binom{n}{r} + \sum_{r=0}^{n} \binom{n}{r} = n \cdot 2^{n-1} + 2^n = (n+2) 2^{n-1}$.

Solution: For even $n$, the maximum binomial coefficient is $\binom{n}{n/2}$. For $n=20$, it is $\binom{20}{10}$.

Solution: For odd $n = 2k+1$, the two middle coefficients $\binom{n}{k}$ and $\binom{n}{k+1}$ are equal and maximal.

Solution: Pascal's identity: $\binom{n}{r} + \binom{n}{r+1} = \frac{n!}{r!(n-r)!} + \frac{n!}{(r+1)!(n-r-1)!} = \frac{(n+1)!}{(r+1)!(n-r)!} = \binom{n+1}{r+1}$.

Solution: Differentiating $(1+x)^n = \sum \binom{n}{r} x^r$ with respect to $x$ gives $n(1+x)^{n-1} = \sum r \binom{n}{r} x^{r-1}$. Setting $x=1$: $n \cdot 2^{n-1} = \sum r \binom{n}{r}$.

Solution: Using differentiation twice: $\sum r^2 \binom{n}{r} = n(n+1) 2^{n-2}$.

Solution: This is Vandermonde's identity: $\sum_{r=0}^{\min(n,m)} \binom{n}{r} \binom{m}{r} = \binom{n+m}{n}$ (coefficient of $x^n$ in $(1+x)^{n+m}$).

Solution: Setting $x = \frac{1}{2}$ in $(1+x)^n$: $\left(1 + \frac{1}{2}\right)^n = \left(\frac{3}{2}\right)^n = \sum_{r=0}^{n} \binom{n}{r} \left(\frac{1}{2}\right)^r$.

Solution: For even $n = 10$, there is one middle term: the $(n/2 + 1)$-th term, which is $\binom{10}{5} x^5$.

Solution: For odd $n = 11$, there are two middle terms at positions $(n+1)/2$ and $(n+3)/2$, i.e., the 6th and 7th terms: $\binom{11}{5} x^5$ and $\binom{11}{6} x^6$.

Solution: The sum is $\frac{(1+x)^{n+1} - x^{n+1}}{1+x-x} = (1+x)^{n+1} - x^{n+1}$. Coefficient of $x^n$ is $\binom{n+1}{n} - 0 = n+1$... Wait, let me recalculate. Actually, sum as a geometric series: the coefficient of $x^n$ comes from $(1+x)^n$ term giving $1$, and from $x^{n-1}(1+x)^{n-1}$ giving $\binom{n-1}{n-1} = 1$, etc. This sums to $2^n - 1$.

Solution: From $\binom{n}{r} = \binom{n}{n-r}$, we have $\binom{n}{3} = \binom{n}{5}$ implies $3 = n - 5$ or $3 + 5 = n$, so $n = 8$.

Solution: From $(1+1)^{20} = 2^{20}$ and $(1-1)^{20} = 0$, adding gives $2 \times (\text{sum of even-positioned}) = 2^{20}$, so even sum $= 2^{19}$.

Solution: The multinomial theorem is a generalization of the binomial theorem. Each term has exponents $n_1, n_2, \ldots, n_k$ that sum to $n$.

Solution: The number of terms in $(x_1 + x_2 + \ldots + x_k)^n$ is $\binom{n+k-1}{k-1} = \binom{10+3-1}{3-1} = \binom{12}{2} = 66$.

Solution: By the multinomial theorem, the coefficient is $\frac{n!}{p! q! r!}$, which is the multinomial coefficient.

Solution: Setting $a = b = c = 1$: $(1+1+1)^5 = 3^5 = 243$.

Solution: This is the coefficient of $x^n$ in $(1+x)^n (1-x)^n = (1-x^2)^n$, which is $0$ for odd $n$ and $(-1)^{n/2} \binom{n}{n/2}$ for even $n$.

Solution: From $\binom{n}{k} = \binom{n}{k+1}$: $\frac{n!}{k!(n-k)!} = \frac{n!}{(k+1)!(n-k-1)!}$, giving $k+1 = n-k$ or $n = 2k+1$. Then $\frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{n-k+1}{k} = \frac{k+2}{k}$. Let me verify: $\frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{(n-k+1)}{k}$. With $n = 2k+1$: $\frac{k+2}{k}$... Actually the direct formula is $\frac{n-k+1}{k}$, which with $n=2k+1$ gives $\frac{k+2}{k}$. But the answer should be in terms of $n,k$. Let me reconsider: we simply have $\frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{n-k+1}{k}$.

Solution: The ratio of consecutive terms in $(x+y)^n$ is $\frac{T_{r+1}}{T_r} = \frac{(n-r+1)y}{rx}$. Setting this equal to $\frac{2(n-r+1)}{r}$ gives $\frac{y}{x} = 2$.