Q.If the seventh terms from the beginning and the end in the expansion of $\left(\sqrt[3]{2} + \dfrac{1}{\sqrt[3]{3}}\right)^{n}$ are equal, then $n$ equals ______ .

Showing the 12 most recent of 25 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.A student is allowed to select atmost '$n$' books from a collection of $2n + 1$ books. If the total number of ways in which he can select atleast one book is $255$, then the value of '$n$' is (A) $6$ (B) $5$ (C) $4$ (D) $3$

   ›Reveal solutionSolution

   $\sum_{r=1}^{n}\binom{2n+1}{r}=2^{2n}-1=255\Rightarrow 2^{2n}=256\Rightarrow n=4$.

   The number of ways to choose at least one and at most $n$ books from $2n+1$ books is

   $N=\binom{2n+1}{1}+\binom{2n+1}{2}+\cdots+\binom{2n+1}{n}.$

   Use the symmetry $\binom{2n+1}{r}=\binom{2n+1}{2n+1-r}$. The full sum of all coefficients is

   $\sum_{r=0}^{2n+1}\binom{2n+1}{r}=2^{2n+1},$

   and it splits into two equal halves ($r=0\ldots n$ and $r=n+1\ldots 2n+1$), so …

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2. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.In a binomial distribution $P(X=2) \div P(X=23) = \left(\dfrac{2}{3}\right)^{21}$, then mean of binomial distribution is (A) 10 (B) 15 (C) 20 (D) 25

   ›Reveal solutionSolution

   A binomial ratio $P(X=2)/P(X=23)=(2/3)^{21}$ pins down both $n$ and $p$; the mean is $np=15$.

   Concept and Intuition

   For $X\sim B(n,p)$, $P(X=r)=\binom{n}{r}p^rq^{n-r}$ with $q=1-p$. Comparing two probabilities at different $r$ values gives a ratio involving both the binomial coefficients and a power of $q/p$. When the given ratio is a pure power of a single fraction (here $(2/3)^{21}$ with no leftover coefficient), the intended reading is that the combinatorial factor cancels out, i.e. $\binom{n}{2}=\binom{n}{23}$.

   Step-by-Step Solution

   1. Write $\dfrac{P(X=2)}{P(X=23)}=\dfrac{\binom{n}{2}p^2q^{n-2}}{\binom{n}{23}p^{23}q^{n-23}}=\dfrac{\binom{n}{2}}{\binom{n}{23}}\left(\dfrac{q}{p}\right)^{21}$.
   2. Since the right side of the given equation is exactly $\left(\dfrac{2}{3}\right)^{21}$ with no extra combinatorial factor, we need $\binom{n}{2}=\binom{n}{23}$.
   3. Using $\binom{n}{r}=\binom{n}{n-r}$, this holds when $n-2=23$, i.e. $n=25$. …

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3. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.All the solutions $(n,r)$ of the equation $\dfrac{{}^nC_r}{{}^{(n+1)}C_r} = \dfrac{1}{3}$ can be obtained from one of the following equations given in the options for $k = 1,2,3,\dots$. Choose the correct option. (A) $\dfrac{{}^{(3k)}C_{2k}}{{}^{(3k+1)}C_{2k}} = \dfrac{1}{3}$ (B) $\dfrac{{}^{(3k-1)}C_{2k}}{{}^{(3k)}C_{2k}} = \dfrac{1}{3}$ (C) $\dfrac{{}^{(4k)}C_{2k}}{{}^{(4k+1)}C_{2k}} = \dfrac{1}{3}$ (D) $\dfrac{{}^{(4k-1)}C_{2k}}{{}^{(4k)}C_{2k}} = \dfrac{1}{3}$

   ›Reveal solutionSolution

   Simplifying the ratio of consecutive binomial coefficients reduces the equation to a simple linear relation between $n,r$; parametrizing by $k$ reproduces option (B) exactly.

   Concept and Intuition

   Ratios of binomial coefficients with the same lower index but consecutive upper indices simplify very cleanly, since most of the factorial terms cancel — this converts what looks like a combinatorial equation into ordinary algebra in $n,r$.

   Step-by-Step Solution

   1. $\dfrac{^nC_r}{^{(n+1)}C_r} = \dfrac{n!/(r!(n-r)!)}{(n+1)!/(r!(n+1-r)!)} = \dfrac{n!\,(n+1-r)!}{(n-r)!\,(n+1)!} = \dfrac{n+1-r}{n+1}$.
   2. Set this equal to $\tfrac13$: $3(n+1-r)=n+1 \Rightarrow 3n+3-3r=n+1 \Rightarrow 2n+2=3r \Rightarrow r=\dfrac{2(n+1)}{3}$.
   3. For $r$ to be an integer, $n+1$ must be divisible by 3. Write $n+1=3k$, i.e. $n=3k-1$, giving $r=2k$.
   4. Substituting back: the family of solutions is exactly $(n,r)=(3k-1,\,2k)$ for $k=1,2,3,\dots$, i.e. $\dfrac{^{(3k-1)}C_{2k}}{^{(3k)}C_{2k}}=\dfrac13$. …

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4. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If $a$ and $b$ are the greatest values of ${}^{2n}C_r$ and ${}^{(2n-1)}C_r$ respectively, then (A) $a = 2b$ (B) $b = 2a$ (C) $a = b$ (D) $a^2 = 2b^2$

   ›Reveal solutionSolution

   This tests where binomial coefficients are maximised for even vs. odd upper index, and the Pascal's-triangle identity linking them. Answer: $a=2b$.

   Concept and Intuition

   For a fixed upper index $N$, the binomial coefficients ${}^{N}C_r$ are largest at the middle. If $N$ is even ($N=2n$), there is a single largest coefficient at $r=n$. If $N$ is odd ($N=2n-1$), the two middle values $r=n-1$ and $r=n$ are tied for the largest (since $(2n-1)-(n-1)=n$, symmetry gives ${}^{2n-1}C_{n-1}={}^{2n-1}C_n$).

   Step-by-Step Solution

   1. Greatest value of ${}^{2n}C_r$: this is $a = {}^{2n}C_n$.
   2. Greatest value of ${}^{(2n-1)}C_r$: this is $b = {}^{2n-1}C_n$ (equal to ${}^{2n-1}C_{n-1}$).
   3. Pascal's rule: ${}^{2n}C_n = {}^{2n-1}C_{n-1} + {}^{2n-1}C_n$. …

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5. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.If $\binom{p}{q}={}^{p}C_q$ and $\displaystyle\sum_{i=0}^{m}\binom{10}{i}\binom{20}{m-i}$ is maximum, then $m=$ (A) $10$ (B) $12$ (C) $15$ (D) $20$

   ›Reveal solutionSolution

   This tests recognizing a Vandermonde convolution and then using the standard fact that binomial coefficients $\binom{n}{r}$ are maximized at the central term. Answer: $m=15$.

   Concept and Intuition

   A sum of the form $\sum_i \binom{p}{i}\binom{q}{m-i}$ is exactly the coefficient of $x^m$ in the product $(1+x)^p(1+x)^q=(1+x)^{p+q}$ — this is Vandermonde's identity. Once the sum collapses to a single binomial coefficient $\binom{p+q}{m}$, the question of "for which $m$ is it maximum" becomes the standard fact that $\binom{n}{r}$ is largest when $r$ is as close as possible to $n/2$ (exactly $n/2$ when $n$ is even).

   Step-by-Step Solution

   1. Recognize $\displaystyle\sum_{i=0}^{m}\binom{10}{i}\binom{20}{m-i}$ as the coefficient of $x^m$ in $(1+x)^{10}\cdot(1+x)^{20}=(1+x)^{30}$.
   2. That coefficient is simply $\binom{30}{m}$. …

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6. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.Let $S_1 = \displaystyle\sum_{j=1}^{10} j(j-1)\cdot{}^{10}C_j$, $S_2 = \displaystyle\sum_{j=1}^{10} j\cdot{}^{10}C_j$ and $S_3 = \displaystyle\sum_{j=1}^{10} j^2\cdot{}^{10}C_j$. Assertion (A): $S_3 = 55\times2^9$. Reason (R): $S_1 = 90\times2^8$ and $S_2 = 10\times2^8$. (A) Both (A) and (R) are true and R is the correct explanation of (A) (B) Both (A) and (R) are true, but (R) is not the correct explanation of (A) (C) (A) is true, but (R) is false (D) (A) is false, but (R) is true

   ›Reveal solutionSolution

   Both $S_1$ and $S_2$ have standard closed forms; checking them against R shows R's $S_1$ claim is correct but its $S_2$ claim is wrong (it should be $10\times2^9$, not $10\times2^8$), making R false overall — while the Assertion, built from the correct values, checks out.

   Concept and Intuition

   The sums $\sum j\binom{n}{j}$ and $\sum j(j-1)\binom{n}{j}$ come from differentiating $(1+x)^n$ once and twice respectively and then setting $x=1$: $\sum j\binom nj = n2^{n-1}$ and $\sum j(j-1)\binom nj=n(n-1)2^{n-2}$. Since $j^2=j(j-1)+j$, the sum of $j^2\binom nj$ is just the sum of these two known results — no need to compute anything from scratch once these standard identities are recalled.

   Step-by-Step Solution

   1. $S_2=\sum_{j=1}^{10} j\binom{10}{j}=10\cdot2^{9}=10\times512=5120$.
   2. $S_1=\sum_{j=1}^{10} j(j-1)\binom{10}{j}=10\cdot9\cdot2^{8}=90\times256=23040=90\times2^8$.
   3. Reason (R) claims $S_1=90\times2^8$ (matches, true) and $S_2=10\times2^8=2560$ (does not match the correct $5120$). Since R asserts both as a package and one part is numerically wrong, R is false.
   4. $S_3=\sum j^2\binom{10}{j}=S_1+S_2=23040+5120=28160$. …

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7. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.If $a_n = \displaystyle\sum_{r=0}^{n} \dfrac{1}{{}^{n}C_r}$ then $\displaystyle\sum_{r=0}^{n} \dfrac{r}{{}^{n}C_r} =$ (A) $(n-1)a_0$ (B) $n.a_n$ (C) $\dfrac{n}{2}.a_n$ (D) $a_{n+1}$

   ›Reveal solutionSolution

   Reflecting the sum $r\to n-r$ and using the symmetry $\binom{n}{n-r}=\binom{n}{r}$ shows $S=\dfrac{n}{2}a_n$ directly, without evaluating either sum in closed form.

   Concept and Intuition

   Sums involving $\binom{n}{r}$ in the denominator, summed over all $r=0$ to $n$, are prime candidates for the "reflect $r\to n-r$" trick, since $\binom{n}{r}$ is symmetric under this substitution — this often relates the sum of $r$-weighted terms directly back to the unweighted sum $a_n$.

   Step-by-Step Solution

   1. Define $S=\displaystyle\sum_{r=0}^{n}\dfrac{r}{\binom{n}{r}}$ — this is exactly the sum we want to find.
   2. Substitute $r\to n-r$ in the summation index (valid since the sum runs over the full symmetric range $0$ to $n$):

   $S = \sum_{r=0}^{n}\frac{n-r}{\binom{n}{n-r}} = \sum_{r=0}^{n}\frac{n-r}{\binom{n}{r}}$

   (using $\binom{n}{n-r}=\binom{n}{r}$). …

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8. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.If the ratio of the terms equidistant from the middle term in the expansion of $(1+x)^{12}$ is $\dfrac{1}{256}$ ($x \in N$) then sum of all the terms of the expansion $(1+x)^{12}$ is (A) $4^{12}$ or $6^{12}$ (B) $3^{12}$ or $5^{12}$ (C) $6^{12}$ or $7^{12}$ (D) $12^{12}$

   ›Reveal solutionSolution

   The key idea is to use the symmetry property of binomial coefficients: terms equidistant from the middle are equal. Setting their ratio to 1/256 leads to an equation in x, which gives two possible values for x. Substituting these into (1+x)^12 yields the sum of all terms, matching option (B).

   We are given the expansion of $(1+x)^{12}$. The middle term is the 7th term (since 12 is even, there is one middle term). Terms equidistant from the middle are symmetric in the sense that their binomial coefficients are equal. For example, the term $r$ places before the middle and the term $r$ places after the middle have coefficients $\binom{12}{6-r}$ and $\binom{12}{6+r}$ respectively. But here the ratio of such terms is given as $\frac{1}{256}$, and $x$ is a natural number. We need to find the sum of all terms, which is $(1+x)^{12}$.

   Why the symmetry property helps:

   In any binomial expansion $(1+x)^n$, the coefficients are symmetric: $\binom{n}{k} = \binom{n}{n-k}$. For terms equidistant from the middle, the powers of $x$ differ, so their ratio involves both coefficients and powers of $x$. Setting that ratio equal to $\frac{1}{256}$ gives an equation that determines $x$.

   Let’s work through it step by step.

   1. Identify the general term and the middle term. The general term in $(1+x)^{12}$ is $T_{k+1} = \binom{12}{k} x^k$. Since $n=12$ is even, the middle term is when $k = 6$:

   $T_7 = \binom{12}{6} x^6.$

   2. Consider two terms equidistant from the middle. Let the term $p$ places before the middle be $T_{7-p}$ and the term $p$ places after the middle be $T_{7+p}$. Their expressions are:

   $T_{7-p} = \binom{12}{6-p} x^{6-p}, \quad T_{7+p} = \binom{12}{6+p} x^{6+p}.$

   3. Form the ratio of these terms. The problem states the ratio of the terms equidistant from the middle is $\frac{1}{256}$. It does not specify which term is numerator, but since $x$ is natural, the term with smaller power of $x$ is smaller if $x>1$. The ratio given is less than 1, so likely the term before the middle (smaller power) over the term after the middle (larger power) equals $\frac{1}{256}$.

   $\frac{T_{7-p}}{T_{7+p}} = \frac{\binom{12}{6-p} x^{6-p}}{\binom{12}{6+p} x^{6+p}} = \frac{\binom{12}{6-p}}{\binom{12}{6+p}} \cdot \frac{1}{x^{2p}}.$

   4. Use symmetry of binomial coefficients. Since $\binom{12}{6-p} = \binom{12}{12-(6-p)} = \binom{12}{6+p}$, the coefficients cancel exactly:

   $\frac{\binom{12}{6-p}}{\binom{12}{6+p}} = 1.$

   So the ratio simplifies to:

   $\frac{1}{x^{2p}} = \frac{1}{256}.$

   Hence $x^{2p} = 256$.

   5. Determine possible values of $p$ and $x$. Since $x$ is a natural number and $p$ is a positive integer (distance from the middle), we have $x^{2p} = 256 = 2^8$.
      • If $p=1$, then $x^2 = 256$ → $x = 16$ (since $x \in \mathbb{N}$).
      • If $p=2$, then $x^4 = 256$ → $x = 4$. …

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9. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.If the coefficients of $(2r+6)^{th}$ and $(r-1)^{th}$ terms in the expansion of $(1+x)^{21}$ are equal, then the value of $r=$ (A) 7 (B) 5 (C) 6 (D) 8

   ›Reveal solutionSolution

   This tests the "equal binomial coefficients" symmetry $\binom{n}{p}=\binom{n}{q}\iff p+q=n$ (or $p=q$); it gives $r=6$.

   Concept and Intuition

   When two binomial coefficients from the same row $\binom{n}{\cdot}$ are equal, either the lower indices are identical, or they are "complementary" and sum to $n$ (since $\binom{n}{p}=\binom{n}{n-p}$). Here the indices coming from term positions $(2r+6)$ and $(r-1)$ can't sensibly be forced equal (that gives a negative $r$), so the complementary case is the one that applies.

   Step-by-Step Solution

   1. The $(2r+6)$-th term of $(1+x)^{21}$ has coefficient $\binom{21}{2r+6-1}=\binom{21}{2r+5}$.
   2. The $(r-1)$-th term has coefficient $\binom{21}{r-1-1}=\binom{21}{r-2}$.
   3. Setting these equal and using $\binom{21}{p}=\binom{21}{q} \iff p+q=21$ (the non-trivial case): $(2r+5)+(r-2)=21$.
   4. $3r+3=21 \Rightarrow 3r=18 \Rightarrow r=6$. …

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10. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
    Q.If $C_j = {}^{n}C_j$, then $C_0 C_r + C_1 C_{r+1} + C_2 C_{r+2} + \ldots + C_{n-r} C_n =$ (A) $\dfrac{(2n)!}{(n-2r)!(n+2r)!}$ (B) $\dfrac{(2n)!}{(n-r)!(n+r)!}$ (C) ${}^{2n}C_r$ (D) ${}^{2n}C_{r+1}$

    ›Reveal solutionSolution

    This sum is a direct application of Vandermonde's convolution identity, giving $\binom{2n}{n-r}=\dfrac{(2n)!}{(n-r)!(n+r)!}$. Answer: (B).

    Concept and Intuition

    Vandermonde's identity: $\sum_k \binom{m}{k}\binom{n}{p-k} = \binom{m+n}{p}$. Recognizing a sum of products of binomial coefficients as this convolution (after using $\binom nk=\binom n{n-k}$) collapses it to one binomial coefficient.

    Step-by-Step Solution

    1. We want $S = \sum_{k=0}^{n-r} C_k C_{k+r}$ where $C_j = \binom nj$.
    2. Use $\binom{n}{k+r} = \binom{n}{n-k-r}$.
    3. So $S = \sum_{k} \binom nk \binom{n}{n-k-r}$, the Vandermonde convolution with $p=n-r$. …

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11. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
    Q.The least value of $n$ such that ${}^{(n-1)}C_6 + {}^{(n-1)}C_7 < {}^{n}C_8$ is (A) $14$ (B) $15$ (C) $16$ (D) $17$

    ›Reveal solutionSolution

    Pascal's identity collapses the left side to $\binom n7$; comparing $\binom n7$ and $\binom n8$ via their ratio gives $n>15$, so the least integer $n$ is $16$.

    Concept and Intuition

    Pascal's rule $\binom{n-1}{k-1}+\binom{n-1}{k}=\binom nk$ is the standard tool whenever consecutive binomial coefficients of the same upper index $(n-1)$ appear added together — here with $k=7$: $\binom{n-1}{6}+\binom{n-1}{7}=\binom n7$.

    Step-by-Step Solution

    1. Apply Pascal's rule with $k=7$: $\binom{n-1}{6}+\binom{n-1}{7}=\binom{n}{7}$.
    2. The inequality $\binom{n-1}{6}+\binom{n-1}{7}<\binom n8$ becomes $\binom n7 < \binom n8$.
    3. Compute the ratio: $\dfrac{\binom n8}{\binom n7} = \dfrac{n!/(8!(n-8)!)}{n!/(7!(n-7)!)} = \dfrac{7!(n-7)!}{8!(n-8)!} = \dfrac{n-7}{8}$.
    4. Require this ratio $>1$: $\dfrac{n-7}{8}>1 \Rightarrow n-7>8 \Rightarrow n>15$. …

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12. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
    Q.The least value of n for which $^{(n-1)}C_2 + {}^{(n-1)}C_3 > {}^{n}C_2$ is (A) $7$ (B) $4$ (C) $5$ (D) $6$

    ›Reveal solutionSolution

    Pascal's rule collapses the left side to ⁿC₃, turning the inequality into ⁿC₃>ⁿC₂, which first becomes strictly true at n=6.

    Concept and Intuition

    Recognising the Pascal's-triangle identity ⁿ⁻¹Cᵣ + ⁿ⁻¹Cᵣ₊₁ = ⁿCᵣ₊₁ immediately simplifies the left-hand side, turning a two-term inequality into a clean single-binomial comparison.

    Step-by-Step Solution

    1. By Pascal's identity: ⁿ⁻¹C₂ + ⁿ⁻¹C₃ = ⁿC₃.
    2. The inequality becomes ⁿC₃ > ⁿC₂.
    3. Compute the ratio: ⁿC₃/ⁿC₂ = [n!/(3!(n−3)!)] / [n!/(2!(n−2)!)] = (n−2)/3.
    4. We need (n−2)/3 > 1 ⟹ n > 5.
    5. Check n=5: ⁵C₃=10, ⁵C₂=10 — equal, not strictly greater, so n=5 fails.
    6. Check n=6: ⁶C₃=20, ⁶C₂=15 — 20>15, holds. …

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