Q.Is $3! + 4! = 7!$?

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:

   $E[X] = 0 \cdot P(X=0) + 1 \cdot P(X=1) + 2 \cdot P(X=2)$

   Substitute the probabilities with $k = \frac13$:

   $P(X=0) = 3\left(\frac13\right)^2 = 3 \cdot \frac19 = \frac13$

   $P(X=1) = 2\left(\frac13\right)^2 = 2 \cdot \frac19 = \frac29$

   $P(X=2) = \left(\frac13\right)^2 + \frac13 = \frac19 + \frac13 = \frac19 + \frac39 = \frac49$

   Then:

   $E[X] = 0 \cdot \frac13 + 1 \cdot \frac29 + 2 \cdot \frac49 = \frac29 + \frac89 = \frac{10}{9}$

   Tip

   You can also compute the mean directly in terms of $k$: $E[X] = 0 + 1\cdot 2k^2 + 2\cdot(k^2+k) = 2k^2 + 2k^2 + 2k = 4k^2 + 2k$. Substituting $k = \frac13$ gives $4\cdot\frac19 + \frac23 = \frac49 + \frac69 = \frac{10}{9}$, same result.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

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

   $B\,G\,B\,G\,B\,G\,B\,G\,B,$

   starting and ending with a boy (a girl-start pattern would need 5 girls). Arrange the 5 distinct boys in the 5 boy-slots ($5!$ ways) and the 4 distinct girls in the 4 girl-slots ($4!$ ways):

   $q=5!\times 4!=120\times 24=2880.$

   Ratio.

   $\frac{q}{p}=\frac{2880}{240}=12.$

   ✓Final answer

   $\dfrac{q}{p}=12$ — option (A).

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:

   $\begin{aligned} S_n &= \frac{1}{5}\Bigg[ \left(\frac{1}{1} - \frac{1}{6}\right) + \left(\frac{1}{6} - \frac{1}{11}\right) + \left(\frac{1}{11} - \frac{1}{16}\right) + \cdots \\ &\qquad\qquad + \left(\frac{1}{5n-4} - \frac{1}{5n+1}\right) \Bigg] \end{aligned}$

   All intermediate terms cancel; only the first and last remain.

   5. Simplify the telescoped sum

   $S_n = \frac{1}{5}\left( \frac{1}{1} - \frac{1}{5n+1} \right) = \frac{1}{5}\left( \frac{5n+1 - 1}{5n+1} \right) = \frac{1}{5} \cdot \frac{5n}{5n+1} = \frac{n}{5n+1}$

   Tip

   A common mistake is forgetting the factor $\frac{1}{5}$ from the difference of denominators. Always check: if $b-a = d$, then $\frac{1}{ab} = \frac{1}{d}\left(\frac{1}{a} - \frac{1}{b}\right)$.

   Watch out

   Do not confuse the pattern: here the first factor in each denominator is $1, 6, 11, \dots$ (starting at 1, adding 5). The $k$-th term is $(5k-4)(5k+1)$, not $(5k+1)(5k+6)$ — that would shift the index.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

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 $r\binom{20}{r}=20\binom{19}{r-1}$,

   $\sum_{r}30\binom{20}{r}\binom{29}{r-1}=\sum_{r}r\binom{20}{r}\binom{30}{r}=20\sum_{t}\binom{19}{t}\binom{30}{30-1-t}=20\binom{49}{29}=20\binom{49}{20}.$

   Since $20\binom{49}{20}=30\binom{49}{19}$ (both equal $\tfrac{600\cdot 49!}{20!\,30!}$), the constant term is

   $\binom{50}{20}+30\binom{49}{19}.$

   ✓Final answer

   The constant term is ${}^{50}C_{20}+30\cdot{}^{49}C_{19}$ — option (D).

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.

   $102000 + 10200 + 1020 + 102 = 113322$

   Tip

   A quick check: the sum of all 24 numbers must be divisible by 3 (since each digit sum 17 is not divisible by 3, but the total sum of all numbers is $24 \times \text{average}$; here average is $113322/24 = 4721.75$, not an integer? Actually 113322 is divisible by 6 but not by 3? Wait: $1+1+3+3+2+2=12$, so it is divisible by 3. The average is $4721.75$, which is fine because the numbers are not all integers? No, they are integers; the average is not necessarily an integer. So no contradiction.)

   Watch out

   A common mistake is to forget that each digit appears equally often in each place, not just overall. Another pitfall: using $4! = 24$ but then incorrectly thinking each digit appears $4! / 4 = 6$ times only in the first position — that’s correct, but some forget to multiply by the place value properly.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

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.

   $\int_0^1 (1+x)^n dx = \left[\frac{(1+x)^{n+1}}{n+1}\right]_0^1 = \frac{2^{n+1} - 1}{n+1}$

   Step 4 — equate.

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

   (Quick check with $n=1$: LHS $= 1 + \frac{1}{2} = \frac{3}{2}$; RHS $= \frac{4-1}{2} = \frac{3}{2}$. ✓)

   ✓Final answer

   The correct option is (C): $\dfrac{2^{n+1}-1}{n+1}$.

   ANSWER: C

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

   $S_{n+1} = 5\cdot 2^{n} + 3n\cdot 2^{n-1} = 2^{n-1}(10 + 3n).$

   For $S_{11}$ the expansion has $n+1 = 11$ terms, so $n = 10$:

   $S_{11} = 2^{9}(10 + 30) = 512\cdot 40 = 20480.$

   ✓Final answer

   $S_{11} = 20480$ — option (C).

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 $r=2$

   $T_3 = \binom{10}{2} x^{0} \cdot (-k)^2 = \binom{10}{2} k^2.$

   Since $\binom{10}{2} = 45$, we have $T_3 = 45 k^2$.

   5. Set this equal to the given constant term $405$

   $45 k^2 = 405 \quad\Rightarrow\quad k^2 = \frac{405}{45} = 9.$

   Hence $k = \pm 3$.

   Watch out

   A common mistake is forgetting the sign: $b = -\frac{k}{x^2}$, so $b^r$ includes $(-k)^r$. For $r=2$, $(-k)^2 = k^2$, so the sign disappears — but for odd $r$ it would matter. Here it’s fine, but always check.

   Tip

   The exponent condition $\frac{10-5r}{2}=0$ gives $r=2$ directly. Notice that $10$ and $5$ share a factor, so the algebra is clean. If the numbers were messier, you’d still solve the linear equation $10-5r=0$.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

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:

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

   $\binom{9}{7} = \binom{9}{2} = 36$, $(-1)^7 = -1$, $3^{-5} = \frac{1}{243}$, $2^{2} = 4$. So

   $2 \cdot 36 \cdot (-1) \cdot \frac{1}{243 \cdot 4} = -\frac{72}{972} = -\frac{2}{27}.$

   Simplify: $-\frac{2}{27} = -\frac{4}{54}$.

   ---

   3. Add the contributions

   The constant term is

   $\frac{7}{18} - \frac{2}{27}.$

   Convert to denominator 54:

   $\frac{7}{18} = \frac{21}{54}, \quad \frac{2}{27} = \frac{4}{54}.$

   So

   $\frac{21}{54} - \frac{4}{54} = \frac{17}{54}.$

   ---

   Watch out

   A common mistake is forgetting that the second factor has a negative sign in $B$ and a fractional coefficient, leading to sign errors or wrong powers of 2 and 3. Also, always check that the exponent equation gives an integer $r$ between 0 and 9.

   Tip

   Notice that only two values of $r$ (6 and 7) survive because the exponents from the first factor are spaced by 2 and 3, and the binomial exponent step is 3. This drastically reduces the work.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

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 $3 + i$:

    $\frac{3}{3 - i} \cdot \frac{3 + i}{3 + i} = \frac{3(3 + i)}{(3)^2 - (i)^2} = \frac{9 + 3i}{9 - (-1)} = \frac{9 + 3i}{10}.$

    5. Match with the options The result $\frac{9+3i}{10}$ is exactly option (D).

    Watch out

    A common mistake is forgetting to multiply numerator and denominator by the conjugate, leaving the answer as $\frac{3}{3-i}$ and then trying to match it to one of the given forms. Always simplify fully.

    Tip

    Notice that the denominator $3 - i$ has magnitude $\sqrt{10}$, so the final denominator 10 is just the square of that magnitude — a neat check.

    ✓Final answer

    The correct option is (D).

    ANSWER: D

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:

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

    Watch out

    A common mistake is to treat this as choosing 30 from 60 people without the equal-participation condition, which gives $^{60}C_{30}$ — option (B). That would allow any mix, like 20 boys, 10 girls, 0 teachers, which violates the condition. The condition "participation of number of boys, girls and teachers is equal" means each group contributes the same count, not that each person is equally likely.

    Tip

    Notice that the answer is a product of three identical binomial coefficients. Whenever a problem says "equal representation" from groups of equal size, you're almost always looking at a power of a single binomial coefficient.

    ✓Final answer

    The correct option is (C): $\dfrac{(20!)^3}{(10!)^6}$.

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.

       Observe that $1001 = \binom{14}{10}$. So we have:

    $\binom{n}{10} a^{10} = \binom{14}{10} \left(\frac{3}{2}\right)^{10}$

    This strongly suggests that we can take $n=14$ and $a = \frac{3}{2}$. Then $na = 14 \cdot \frac{3}{2} = 21$.

    But is this the only possibility? Could $n$ be something else? For instance, if $n=15$, then $\binom{15}{10} = 3003$, and we would need $a^{10} = \frac{1001}{3003} \left(\frac{3}{2}\right)^{10} = \frac{1}{3} \left(\frac{3}{2}\right)^{10}$, which gives $a = \frac{3}{2} \cdot \frac{1}{\sqrt[10]{3}}$, not a nice number. The problem likely expects integer or simple rational values for $n$ and $a$ (since $na$ is an integer option), so $n=14$, $a=\frac{3}{2}$ is the natural match.

    Tip

    When matching coefficients of the form $\binom{n}{r} a^r$ to a known number, look for the binomial coefficient to match exactly — here $\binom{14}{10}=1001$ — and then the power of $a$ matches the remaining factor. This avoids solving messy equations.

    6. Compute $na$. With $n=14$ and $a=\frac{3}{2}$, we get:

    $na = 14 \times \frac{3}{2} = 21$

    Watch out

    A common mistake is to forget the $(-1)^r$ factor when expanding a negative exponent. Here $r=10$ is even, so it doesn't affect the sign, but for odd powers it would. Always check the parity.

    ✓Final answer

    The value is $\boxed{21}$, which corresponds to option (D).