Q.If $\dfrac{1}{8!} + \dfrac{1}{9!} = \dfrac{x}{10!}$, find $x$.

Showing the 12 most recent of 16 on this concept.

1. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.The probability distribution of a random variable X is given below. Then the mean of X is

   [!FORMULA] $\begin{array}{c|ccc} X = x_i & 0 & 1 & 2 \\ \hline P(X = x_i) & 3k^2 & 2k^2 & k^2 + k \\ \end{array}$

   (A) $\dfrac{7}{9}$ (B) $\dfrac{5}{9}$ (C) $\dfrac{10}{9}$ (D) $\dfrac{13}{9}$

   ›Reveal solutionSolution

   The key idea is to first find $k$ using the fact that total probability equals 1, then compute the mean as $\sum x_i P(X=x_i)$. The mean is $\frac{10}{9}$, so the correct option is (C).

   We are given a discrete probability distribution for $X$ with values 0, 1, 2 and probabilities expressed in terms of $k$. To find the mean (expected value), we must first determine $k$ from the condition that the sum of all probabilities is 1. Then we compute $E[X] = \sum x_i P(X=x_i)$.

   1. Set up the total probability condition. The probabilities must sum to 1:

   $3k^2 + 2k^2 + (k^2 + k) = 1$

   Simplify:

   $3k^2 + 2k^2 + k^2 + k = 6k^2 + k = 1$

   So we have the quadratic equation:

   $6k^2 + k - 1 = 0$

   2. Solve for $k$. Factor or use the quadratic formula:

   $k = \frac{-1 \pm \sqrt{1 + 24}}{12} = \frac{-1 \pm 5}{12}$

   This gives two candidates:

   $k = \frac{4}{12} = \frac{1}{3} \quad \text{or} \quad k = \frac{-6}{12} = -\frac{1}{2}$

   Since probabilities cannot be negative, we check:

   • For $k = -\frac12$, $3k^2 = 3 \cdot \frac14 = \frac34$ (positive), $2k^2 = \frac12$ (positive), but $k^2 + k = \frac14 - \frac12 = -\frac14$ — negative, which is impossible.
   • For $k = \frac13$, all probabilities are non-negative: $3 \cdot \frac19 = \frac13$, $2 \cdot \frac19 = \frac29$, $\frac19 + \frac13 = \frac49$. Sum = $\frac13 + \frac29 + \frac49 = 1$. Hence $k = \frac13$ is the only valid value.

   Watch out

   A common mistake is to forget that probabilities must be non-negative. Always check each probability after solving for $k$.

   3. Compute the mean $E[X]$. The mean is: $$$$

   🔒 Start your 14-day free trial to see the full solution →
2. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.Let $p$ be the number of ways of arranging 6 students such that 3 are around a circular table and the remaining 3 in a row. Let $q$ be the number of ways of arranging 5 boys and 4 girls in a row such that no two boys and no two girls are together. Then $\dfrac{q}{p} =$ (A) $12$ (B) $18$ (C) $6$ (D) $8$

   ›Reveal solutionSolution

   $p=\binom{6}{3}\cdot(3-1)!\cdot 3!=240$ and $q=5!\cdot 4!=2880$, so $\dfrac{q}{p}=12$ — option (A).

   Compute $p$ (arranging 6 students: 3 around a circle, 3 in a row).

   From the 6 distinct students we first choose which 3 sit at the circular table: $\binom{6}{3}=20$ ways. Those 3 are arranged around the table in $(3-1)!=2!=2$ ways (a circular arrangement fixes one seat to remove rotational symmetry). The remaining 3 are arranged in a row in $3!=6$ ways. Since the three steps are independent,

   $p=\binom{6}{3}\cdot(3-1)!\cdot 3!=20\times 2\times 6=240.$

   Compute $q$ (5 boys, 4 girls in a row, no two boys and no two girls adjacent).

   "No two boys and no two girls together" forces a strictly alternating row. With 5 boys and 4 girls the only alternating pattern is …

   🔒 Start your 14-day free trial to see the full solution →
3. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.$\dfrac{1}{1.6} + \dfrac{1}{6.11} + \dfrac{1}{11.16} + \ldots$ n terms = (A) $\dfrac{n}{5n+1}$ (B) $\dfrac{5}{5n+1}$ (C) $\dfrac{5n}{5n+1}$ (D) $\dfrac{n}{5(5n+1)}$

   ›Reveal solutionSolution

   This sum telescopes after rewriting each term as a difference of fractions using the pattern of denominators. The sum simplifies to $\frac{n}{5n+1}$, which matches option (A).

   We are summing a series of fractions whose denominators are products of numbers in an arithmetic progression:

   $1.6$, $6.11$, $11.16$, …

   Notice the first factor in each denominator increases by 5 each time: 1, 6, 11, 16, …

   So the $k$-th term has denominator $(5k-4)(5k+1)$.

   The key idea: when a fraction has the form $\frac{1}{a \cdot b}$ where $b - a$ is constant, we can rewrite it as $\frac{1}{b-a}\left(\frac{1}{a} - \frac{1}{b}\right)$. This creates a telescoping series — most terms cancel when summed.

   1. Identify the general term The $k$-th term (starting from $k=1$) is:

   $T_k = \frac{1}{(5k-4)(5k+1)}$

   Check: $k=1$ gives $1\cdot6$, $k=2$ gives $6\cdot11$, etc.

   2. Decompose into partial fractions Since $(5k+1) - (5k-4) = 5$, we have:

   $\frac{1}{(5k-4)(5k+1)} = \frac{1}{5}\left(\frac{1}{5k-4} - \frac{1}{5k+1}\right)$

   This is the crucial step — it turns each term into a difference.

   3. Write the sum of $n$ terms

   $S_n = \sum_{k=1}^{n} \frac{1}{(5k-4)(5k+1)} = \frac{1}{5} \sum_{k=1}^{n} \left( \frac{1}{5k-4} - \frac{1}{5k+1} \right)$

   4. Observe the telescoping Write out the first few terms: $\begin{aligned} S_n &= \frac{1}{5}\Bigg[ \left(\frac{1}{1} - \frac{1}{6}\right) + \left(\frac{1}{6} - \frac{1}{11}\right) + \left(\frac{1}{11} - \frac{1}{16}\right) + \cdots \\ &\qquad\qquad + \left(\frac{1}{5n-4} - \frac{1}{5n+1}\right) \Bigg] \end{aligned}$ …

   🔒 Start your 14-day free trial to see the full solution →
4. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.The constant term in the expansion of $\left(1+\frac{1}{x}\right)^{20}\left(30x(1+x)^{29}+(1+x)^{30}\right)$ is (A) ${}^{50}C_{20} + 30 \cdot {}^{50}C_{29}$ (B) ${}^{50}C_{19} + 30 \cdot {}^{49}C_{19}$ (C) ${}^{50}C_{20} + 30 \cdot {}^{49}C_{20}$ (D) ${}^{50}C_{20} + 30 \cdot {}^{49}C_{19}$

   ›Reveal solutionSolution

   The constant term equals ${}^{50}C_{20}+30\,{}^{49}C_{19}$, option (D), obtained by Vandermonde convolution.

   Write the product as $\left(1+\tfrac1x\right)^{20}\big[30x(1+x)^{29}+(1+x)^{30}\big]$. Since $\left(1+\tfrac1x\right)^{20}=\sum_{r}\binom{20}{r}x^{-r}$, the constant term picks the coefficient of $x^{r}$ from the second bracket for each $r$.

   From $(1+x)^{30}$: coefficient of $x^{r}$ is $\binom{30}{r}$, contributing

   $\sum_{r}\binom{20}{r}\binom{30}{r}=\sum_{r}\binom{20}{r}\binom{30}{30-r}=\binom{50}{30}=\binom{50}{20}.$

   From $30x(1+x)^{29}$: coefficient of $x^{r}$ is $30\binom{29}{r-1}$. Using $\binom{29}{r-1}=\tfrac{r}{30}\binom{30}{r}$ and $r\binom{20}{r}=20\binom{19}{r-1}$, …

   🔒 Start your 14-day free trial to see the full solution →
5. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.The sum of all the 4-digit numbers formed by taking all the digits from 2, 3, 5, 7 without repetition, is (A) 331122 (B) 123312 (C) 113322 (D) 132132

   ›Reveal solutionSolution

   The sum of all distinct 4-digit numbers formed from digits 2, 3, 5, 7 without repetition is found by noting each digit appears equally often in each place value; the total is $113322$, which corresponds to option (C).

   Concept and intuition

   When we form all 4-digit numbers using each of the digits 2, 3, 5, 7 exactly once, we are essentially listing every permutation of these four digits. There are $4! = 24$ such numbers. Instead of writing them all out and adding, we use symmetry: each digit appears in the thousands place the same number of times, and similarly for hundreds, tens, and units. So we can compute the total sum by multiplying the sum of the digits by the place-value weight and the number of times each digit occupies that place.

   Step-by-step reasoning

   1. Count how many numbers and how often each digit appears in each position.

      There are $4! = 24$ distinct 4-digit numbers. In any fixed position (say the thousands place), each of the four digits appears equally often. Since there are 24 numbers and 4 digits, each digit appears $24/4 = 6$ times in each position.

   2. Sum of contributions from one place value.

      The sum of the digits is $2 + 3 + 5 + 7 = 17$. In a given position, each digit appears 6 times, so the total sum contributed by that position (ignoring place value) is $6 \times 17 = 102$.

   3. Apply place values.

      • Thousands place: each digit’s contribution is multiplied by $1000$. So total from thousands = $102 \times 1000 = 102000$.
      • Hundreds place: multiply by $100$ → $102 \times 100 = 10200$.
      • Tens place: multiply by $10$ → $102 \times 10 = 1020$.
      • Units place: multiply by $1$ → $102 \times 1 = 102$.

   4. Add all contributions.

      $102000 + 10200 + 1020 + 102 = 113322$ …

   🔒 Start your 14-day free trial to see the full solution →
6. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.If $(1+x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n$ for $n \in \mathbb{N}$, then $C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \ldots + \frac{C_n}{n+1} =$ (A) $\dfrac{2^n - 1}{n+1}$ (B) $\dfrac{2^n - 1}{n}$ (C) $\dfrac{2^{n+1} - 1}{n+1}$ (D) $\dfrac{2^{n+1} - 1}{n}$

   ›Reveal solutionSolution

   Integrating $(1+x)^n = \sum C_k x^k$ over $[0,1]$ gives $C_0 + \frac{C_1}{2} + \cdots + \frac{C_n}{n+1} = \frac{2^{n+1}-1}{n+1}$.

   Concept — evaluating binomial sums by integration. When each coefficient $C_k$ is divided by $k+1$, the sum is produced by integrating the expansion term by term, since $\int_0^1 x^k dx = \frac{1}{k+1}$.

   Step 1 — integrate both sides of the expansion.

   $\int_0^1 (1+x)^n\,dx = \int_0^1 \left(C_0 + C_1x + C_2x^2 + \cdots + C_nx^n\right)dx$

   Step 2 — right side (term by term).

   $C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \cdots + \frac{C_n}{n+1}$

   Step 3 — left side. …

   🔒 Start your 14-day free trial to see the full solution →
7. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
   Q.Let $c_0, c_1, c_2, \ldots, c_n$ be the binomial coefficients in the expansion of $(1 + x)^n$. If $S_{n+1} = 5.c_0 + 8.c_1 + 11.c_2 + \ldots$ (n + 1) terms, then $S_{11} =$ (A) 18944 (B) 17920 (C) 20480 (D) 40960

   ›Reveal solutionSolution

   Writing the coefficients as $5+3k$ gives $S_{n+1} = 2^{n-1}(3n+10)$, so $S_{11} = 2^{9}\cdot 40 = 20480$ (option C).

   The multipliers $5,8,11,\ldots$ are $5+3k$ for $k = 0,1,\ldots,n$, so

   $S_{n+1} = \sum_{k=0}^{n}(5+3k)\,c_k = 5\sum_{k=0}^{n}c_k + 3\sum_{k=0}^{n}k\,c_k.$

   Using the standard sums for $(1+x)^n$:

   $\sum_{k=0}^{n} c_k = 2^{n}, \qquad \sum_{k=0}^{n} k\,c_k = n\,2^{n-1},$

   we get …

   🔒 Start your 14-day free trial to see the full solution →
8. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
   Q.If the term independent of $x$ in the expansion of $\left(\sqrt{x} - \dfrac{k}{x^2}\right)^{10}$ is $405$, then $k=$ (A) $\pm 1$ (B) $0$ (C) $\pm 3$ (D) $\pm 5$

   ›Reveal solutionSolution

   The term independent of $x$ occurs when the powers of $x$ cancel. Using the binomial theorem, we find $k^2 = 9$, so $k = \pm 3$, which corresponds to option (C).

   We are asked: for what value(s) of $k$ does the constant term (independent of $x$) in the expansion of $\left(\sqrt{x} - \frac{k}{x^2}\right)^{10}$ equal $405$? The key is to recall that in a binomial expansion, each term has a specific power of $x$, and we want the one where that power is zero.

   Concept and intuition:

   The binomial theorem tells us that

   $(a+b)^n = \sum_{r=0}^n \binom{n}{r} a^{n-r} b^r.$

   Here $a = \sqrt{x} = x^{1/2}$ and $b = -\frac{k}{x^2} = -k x^{-2}$. So the general term is

   $T_r = \binom{10}{r} (x^{1/2})^{10-r} \cdot (-k x^{-2})^r.$

   The exponent of $x$ in $T_r$ is $\frac{1}{2}(10-r) - 2r$. We set this equal to $0$ to find the term independent of $x$. Then we equate that term’s coefficient to $405$ and solve for $k$.

   Step-by-step solution:

   1. Write the general term

   $T_{r+1} = \binom{10}{r} (x^{1/2})^{10-r} \left(-\frac{k}{x^2}\right)^r = \binom{10}{r} x^{\frac{10-r}{2}} \cdot (-k)^r x^{-2r}.$

   2. Combine the powers of $x$

   $x^{\frac{10-r}{2} - 2r} = x^{\frac{10-r - 4r}{2}} = x^{\frac{10 - 5r}{2}}.$

   3. Set the exponent to zero (for the term independent of $x$)

   $\frac{10 - 5r}{2} = 0 \quad\Rightarrow\quad 10 - 5r = 0 \quad\Rightarrow\quad r = 2.$

   4. Find the term for $r=2$ $T_3 = \binom{10}{2} x^{0} \cdot (-k)^2 = \binom{10}{2} k^2.$ …

   🔒 Start your 14-day free trial to see the full solution →
9. TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
   Q.The term independent of $x$ in the expansion of $(1-3x+2x^3)\left(\dfrac{3x^2}{2}-\dfrac{1}{3x}\right)^9$ is (A) $\dfrac{7}{18}$ (B) $\dfrac{5}{18}$ (C) $\dfrac{19}{54}$ (D) $\dfrac{17}{54}$

   ›Reveal solutionSolution

   The key idea is to expand the second factor using the binomial theorem, then multiply by the first factor and collect terms where the total exponent of $x$ is zero. The constant term is $\frac{17}{54}$, so the correct option is (D).

   We need the term independent of $x$ in

   $(1-3x+2x^3)\left(\frac{3x^2}{2} - \frac{1}{3x}\right)^9.$

   The second factor is a binomial with a fractional coefficient and a negative power of $x$. The constant term will come from multiplying each term of the first factor by a term from the expansion of the second factor such that the total power of $x$ cancels.

   ---

   1. Expand the second factor using the binomial theorem

   Let

   $A = \frac{3x^2}{2}, \quad B = -\frac{1}{3x}.$

   Then

   $\left(A + B\right)^9 = \sum_{r=0}^{9} \binom{9}{r} A^{9-r} B^r.$

   Substitute back:

   $A^{9-r} = \left(\frac{3x^2}{2}\right)^{9-r} = \frac{3^{9-r}}{2^{9-r}} x^{2(9-r)},$

   $B^r = \left(-\frac{1}{3x}\right)^r = (-1)^r \frac{1}{3^r} x^{-r}.$

   So the general term is

   $T_r = \binom{9}{r} \frac{3^{9-r}}{2^{9-r}} (-1)^r \frac{1}{3^r} \, x^{2(9-r) - r}.$

   Simplify the coefficient:

   $\frac{3^{9-r}}{3^r} = 3^{9-2r},$

   so

   $T_r = \binom{9}{r} (-1)^r \frac{3^{9-2r}}{2^{9-r}} \, x^{18 - 3r}.$

   Thus the exponent of $x$ in $T_r$ is $18 - 3r$.

   ---

   2. Multiply by each term of the first factor

   The first factor is $1 - 3x + 2x^3$. We multiply each term by $T_r$ and look for total exponent zero.

   • From $1 \cdot T_r$: exponent is $18 - 3r$. Set $18 - 3r = 0 \Rightarrow r = 6$. Term:

   $\binom{9}{6} (-1)^6 \frac{3^{9-12}}{2^{9-6}} = \binom{9}{3} \cdot 1 \cdot \frac{3^{-3}}{2^{3}}.$

   $\binom{9}{3} = 84$, $3^{-3} = \frac{1}{27}$, $2^3 = 8$. So this term is

   $84 \cdot \frac{1}{27 \cdot 8} = \frac{84}{216} = \frac{7}{18}.$

   • From $(-3x) \cdot T_r$: exponent is $1 + (18 - 3r) = 19 - 3r$. Set $19 - 3r = 0 \Rightarrow r = 19/3$, not an integer → no contribution.

   • From $(2x^3) \cdot T_r$: exponent is $3 + (18 - 3r) = 21 - 3r$. Set $21 - 3r = 0 \Rightarrow r = 7$.

     Term:

     $2 \cdot \binom{9}{7} (-1)^7 \frac{3^{9-14}}{2^{9-7}}.$ …

   🔒 Start your 14-day free trial to see the full solution →
10. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
    Q.If $i = \sqrt{-1}$ then $\sum_{n=0}^{\infty} \left(\frac{i}{3}\right)^n$ (A) $\frac{9-3i}{10}$ (B) $9-3i$ (C) $9+3i$ (D) $\frac{9+3i}{10}$

    ›Reveal solutionSolution

    This is a geometric series with ratio $i/3$; its sum is $\frac{1}{1 - i/3} = \frac{9+3i}{10}$, which matches option (D).

    The key idea is recognizing that the infinite sum $\sum_{n=0}^{\infty} r^n$ converges to $\frac{1}{1-r}$ when $|r| < 1$. Here $r = \frac{i}{3}$, and $|i/3| = 1/3 < 1$, so the formula applies directly. The only twist is simplifying the complex fraction to match one of the given choices.

    1. Identify the series type

       The sum is $\sum_{n=0}^{\infty} \left(\frac{i}{3}\right)^n$. This is a geometric series with first term $1$ (when $n=0$) and common ratio $r = \frac{i}{3}$.

    2. Check convergence

       The magnitude of the ratio is $|r| = \left|\frac{i}{3}\right| = \frac{1}{3} < 1$, so the series converges absolutely.

    3. Apply the geometric series formula

       For $|r| < 1$,

    $\sum_{n=0}^{\infty} r^n = \frac{1}{1 - r}.$

    Substituting $r = \frac{i}{3}$:

    $\sum_{n=0}^{\infty} \left(\frac{i}{3}\right)^n = \frac{1}{1 - \frac{i}{3}}.$

    4. Simplify the complex fraction Multiply numerator and denominator by the conjugate of the denominator (or simply by 3):

    $\frac{1}{1 - \frac{i}{3}} = \frac{1}{\frac{3 - i}{3}} = \frac{3}{3 - i}.$

    Now rationalize by multiplying numerator and denominator by $3 + i$:

    $$$$

    🔒 Start your 14-day free trial to see the full solution →
11. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
    Q.The number of ways of selecting a committee of 30 persons from 20 boys, 20 girls and 20 teachers such that the participation of number of boys, girls and teachers in that committee is equal, is (A) $(20!)(20!)(20!)$ (B) $^{60}C_{30}$ (C) $\dfrac{(20!)^3}{(10!)^6}$ (D) $\dfrac{(20!)(20!)}{10! \ 10!}$

    ›Reveal solutionSolution

    The key idea is that equal participation means exactly 10 from each group, so the answer is the product of three independent selections: $\binom{20}{10}^3$, which simplifies to $\frac{(20!)^3}{(10!)^6}$ — option (C).

    The problem asks for a committee of 30 persons from three groups — 20 boys, 20 girls, 20 teachers — with the condition that the number from each group is equal. That means the committee must have the same count of boys, girls, and teachers. Since 30 divided equally among three groups gives 10 each, the committee must consist of exactly 10 boys, 10 girls, and 10 teachers.

    The selection from each group is independent of the others. For the boys, we choose 10 out of 20; for the girls, 10 out of 20; for the teachers, 10 out of 20. The total number of ways is the product of these three independent choices.

    1. Number of ways to choose 10 boys from 20:

       This is $\binom{20}{10} = \frac{20!}{10! \, 10!}$.

    2. Number of ways to choose 10 girls from 20:

       Similarly, $\binom{20}{10} = \frac{20!}{10! \, 10!}$.

    3. Number of ways to choose 10 teachers from 20:

       Again, $\binom{20}{10} = \frac{20!}{10! \, 10!}$.

    Since the selections are independent, multiply them:

    $\text{Total ways} = \binom{20}{10} \times \binom{20}{10} \times \binom{20}{10} = \left( \frac{20!}{10! \, 10!} \right)^3 = \frac{(20!)^3}{(10!)^6}.$ …

    🔒 Start your 14-day free trial to see the full solution →
12. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
    Q.In the expansion of $\left(1+\dfrac{3x}{2}\right)^{-5}$, the coefficient of $x^{10}$ is equal to the coefficient of $x^{10}$ in $(1+ax)^n, n \in \mathbb{N}$, then $na =$ (A) 15 (B) 18 (C) 24 (D) 21

    ›Reveal solutionSolution

    The key idea is to expand the given binomial with a negative integer exponent using the general binomial theorem, match the coefficient of $x^{10}$ to the form $(1+ax)^n$, and then compute $na$. The final value is $21$.

    The problem asks us to compare the coefficient of $x^{10}$ in two expansions. The first is $\left(1+\frac{3x}{2}\right)^{-5}$, which is a binomial with a negative integer exponent. The second is $(1+ax)^n$, where $n$ is a natural number — a standard positive integer exponent. The condition says these coefficients are equal, and we need $na$.

    The core concept here is the general binomial theorem. For a positive integer exponent $n$, we have the familiar expansion:

    $(1+ax)^n = \sum_{r=0}^{n} \binom{n}{r} (ax)^r$

    But for a negative integer exponent, say $(1+u)^{-m}$ where $m$ is a positive integer, the expansion is an infinite series:

    $(1+u)^{-m} = \sum_{r=0}^{\infty} \binom{-m}{r} u^r$

    where the binomial coefficient for a negative integer is defined as:

    $\binom{-m}{r} = \frac{(-m)(-m-1)(-m-2)\cdots(-m-r+1)}{r!} = (-1)^r \binom{m+r-1}{r}$

    This last form is crucial — it turns the negative sign into a manageable expression.

    Let’s work through it step by step.

    1. Expand the given expression. We have $\left(1+\frac{3x}{2}\right)^{-5}$. Here $u = \frac{3x}{2}$ and $m=5$. The general term in the expansion is:

    $T_r = \binom{-5}{r} \left(\frac{3x}{2}\right)^r$

    Using the formula $\binom{-5}{r} = (-1)^r \binom{5+r-1}{r} = (-1)^r \binom{r+4}{r}$, we get:

    $T_r = (-1)^r \binom{r+4}{r} \left(\frac{3}{2}\right)^r x^r$

    2. Extract the coefficient of $x^{10}$. For $x^{10}$, set $r=10$. The coefficient is:

    $C_1 = (-1)^{10} \binom{10+4}{10} \left(\frac{3}{2}\right)^{10} = \binom{14}{10} \left(\frac{3}{2}\right)^{10}$

    Since $\binom{14}{10} = \binom{14}{4} = \frac{14 \cdot 13 \cdot 12 \cdot 11}{4 \cdot 3 \cdot 2 \cdot 1} = 1001$, we have:

    $C_1 = 1001 \cdot \left(\frac{3}{2}\right)^{10}$

    3. Now consider the second expansion. For $(1+ax)^n$ with $n \in \mathbb{N}$, the coefficient of $x^{10}$ comes from the term where $r=10$ (provided $n \geq 10$, otherwise the coefficient is zero — but since it must equal a non-zero coefficient, $n \geq 10$). The coefficient is:

    $C_2 = \binom{n}{10} a^{10}$

    4. Set the coefficients equal. The problem states $C_1 = C_2$, so:

    $1001 \cdot \left(\frac{3}{2}\right)^{10} = \binom{n}{10} a^{10}$

    This is one equation in two unknowns ($n$ and $a$), but we are not asked for them individually — only $na$. We need to find a way to match the structure.

    5. Match the form cleverly. Notice that $\binom{n}{10} a^{10}$ can be written as $\left( \sqrt[10]{\binom{n}{10}} \cdot a \right)^{10}$. But a more elegant approach: the coefficient in the first expansion came from a binomial with exponent $-5$, which suggests that the second expansion might be related by choosing $n$ and $a$ such that the binomial coefficient and the power of $a$ combine to give the same numeric factor. …

    🔒 Start your 14-day free trial to see the full solution →