Q.Three letters can be posted in five letterboxes in $3^5$ ways.

Showing the 12 most recent of 25 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If the number of circular permutations of 10 distinct things taken 5 at a time is m and the number of linear permutations of 9 distinct things taken 4 at a time is n, then m : n = (A) 1 : 2 (B) 2 : 1 (C) 2 : 3 (D) 3 : 2

   ›Reveal solutionSolution

   The key idea is to correctly apply the formulas for circular and linear permutations, then compute the ratio. The final ratio is $2 : 1$.

   The problem asks for the ratio of two counts: $m$, the number of circular permutations of 10 distinct things taken 5 at a time, and $n$, the number of linear permutations of 9 distinct things taken 4 at a time. The trick is that circular permutations are not the same as linear ones — in a circle, rotations are considered identical, so we divide by the number of positions.

   Let’s break it down.

   1. Understanding circular permutations

      When arranging $r$ distinct objects around a circle, there is no fixed starting point. A rotation of the entire arrangement gives the same circular arrangement. So the number of circular permutations of $r$ distinct objects is $\frac{r!}{r} = (r-1)!$.

      But here we are selecting 5 out of 10 distinct things first, then arranging them in a circle. So we first choose which 5 things to use, then arrange them circularly.

      Number of ways to choose 5 things from 10: $\binom{10}{5}$.

      Number of circular arrangements of these 5 distinct things: $(5-1)! = 4!$.

      Hence,

   $m = \binom{10}{5} \times 4!$

   2. Simplify $m$

   $\binom{10}{5} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$

   $4! = 24$

   So $m = 252 \times 24 = 6048$.

   3. Understanding linear permutations Linear permutations of $n$ distinct things taken $r$ at a time is simply $P(n, r) = \frac{n!}{(n-r)!}$. Here $n=9$, $r=4$. …

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2. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.Let $S = \{2, 3, 5, 7, 11, 13\}$ be a set. Consider the set of all possible onto functions from $S$ to $S$. If a function $f$ is chosen from that set, then the probability that it satisfies the condition $f(3) > 3f(2)$ is (A) $\dfrac{2}{3}$ (B) $\dfrac{2}{15}$ (C) $\dfrac{1}{6}$ (D) $\dfrac{1}{10}$

   ›Reveal solutionSolution

   Onto functions on a 6-element set are the $6!$ permutations. Only $5$ ordered value-pairs $(f(2),f(3))$ satisfy $f(3) > 3f(2)$, giving probability $\tfrac{5}{30} = \tfrac16$.

   Onto = bijection. A function from the 6-element set $S$ onto itself is a permutation, so there are $6! = 720$ of them (the sample space).

   Count the favourable case. Fix the images $f(2) = a$ and $f(3) = b$ (distinct elements of $S$) with $b > 3a$; the remaining $4$ elements can map to the remaining $4$ values in $4! = 24$ ways. Count pairs $(a,b)$ with $a \ne b$ and $b > 3a$:

   • $a = 2\;(3a=6)$: $b \in \{7, 11, 13\}$ — $3$ pairs …

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3. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.There are 7 men and 5 women in a park. The number of ways of arranging them around a circular path such that 4 particular persons which include 2 particular men and 2 particular women never stand together is (A) $11879(8!)$ (B) $966(8!)$ (C) $986(8!)$ (D) $494(4!)(8!)$

   ›Reveal solutionSolution

   The key idea is to count total circular arrangements of 12 people, then subtract those where the 4 particular persons (2 men, 2 women) are forced to sit together as a block, using the circular permutation formula $(n-1)!$ and treating the block as a single entity. The final answer is $966(8!)$, which corresponds to option (B).

   ---

   Concept and Intuition

   When arranging people around a circle, the number of distinct arrangements is $(n-1)!$ because rotations are considered identical. Here, we have 12 people total. The problem asks for arrangements where 4 particular persons (two specific men and two specific women) are never all together. The classic trick: count all arrangements, then subtract the arrangements where they are together. The "together" case is handled by treating the 4 people as a single block, then arranging the block + the remaining 8 people around the circle. But careful: inside the block, the 4 people can be permuted among themselves. Also, the block itself can be placed anywhere in the circle. Let's work step by step.

   ---

   Step-by-step solution

   1. Total circular arrangements of all 12 people

      For $n$ distinct people around a circle, the number is $(n-1)!$. Here $n=12$, so total arrangements = $(12-1)! = 11!$.

   2. Count arrangements where the 4 particular persons are always together

      Treat the 4 particular persons as a single "block". Then we have this block plus the other 8 individuals, making $1 + 8 = 9$ entities to arrange around the circle.

      Number of circular arrangements of these 9 entities = $(9-1)! = 8!$.

   3. Internal arrangements of the block

      Inside the block, the 4 particular persons can be arranged in $4!$ ways. So the total number of arrangements where they are together = $8! \times 4!$.

   4. Subtract to get "never together" …

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4. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.Bag A contains 5 white and 2 black balls. Bag B contains 2 white and 5 black balls. Two balls are randomly chosen from bag A and placed in bag B. Now a ball is drawn randomly from bag B and found that it is white. The probability that the two balls drawn from bag A are of different colour is (A) $\dfrac{10}{21}$ (B) $\dfrac{10}{63}$ (C) $\dfrac{5}{12}$ (D) $\dfrac{2}{189}$

   ›Reveal solutionSolution

   Set up the three transfer cases (WW, BB, WB) from bag A, find the probability of drawing white from the enlarged bag B, then apply Bayes' theorem; the answer is $\dfrac{5}{12}$.

   Bag A has $5$ white, $2$ black. Choosing $2$ balls, with $\binom{7}{2}=21$:

   $P(\text{WW})=\frac{\binom52}{21}=\frac{10}{21},\quad P(\text{BB})=\frac{\binom22}{21}=\frac{1}{21},\quad P(\text{WB})=\frac{5\cdot2}{21}=\frac{10}{21}.$

   Bag B starts with $2$ white, $5$ black. After adding $2$ balls it holds $9$ balls, and the chance of then drawing white is:

   $P(W\mid\text{WW})=\frac{4}{9},\quad P(W\mid\text{BB})=\frac{2}{9},\quad P(W\mid\text{WB})=\frac{3}{9}.$

   Total probability of drawing white: …

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5. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.A question paper contains 3 parts A, B and C. There are 5 questions in A, 4 questions in B and 3 questions in C. Atleast 3 questions from A, atleast 2 questions from B and atleast 1 question from C must be attempted. If a student has attempted 7 questions with these conditions, then the total number of ways in which he can attempt the questions (A) 330 (B) 402 (C) 390 (D) 430

   ›Reveal solutionSolution

   The problem asks for the number of ways to choose exactly 7 questions from three parts (A:5, B:4, C:3) with minimums of 3 from A, 2 from B, and 1 from C. The answer is found by summing over all valid triples (a,b,c) that sum to 7, using combinations: total = 390, so option (C).

   We need to count the number of ways a student can select questions to attempt, given:

   • Part A: 5 questions, must attempt at least 3.
   • Part B: 4 questions, must attempt at least 2.
   • Part C: 3 questions, must attempt at least 1.
   • Total attempted = exactly 7 questions.

   The key idea: For each part, the student chooses some number of questions from that part, subject to the minimums and the total. Since the questions within each part are distinct, the number of ways to choose $a$ questions from A is $\binom{5}{a}$, similarly $\binom{4}{b}$ and $\binom{3}{c}$. We sum over all triples $(a,b,c)$ that satisfy the constraints.

   ---

   Step-by-step reasoning:

   1. Define variables and constraints

      Let $a$ = number chosen from A, $b$ from B, $c$ from C.

      Constraints:

      • $3 \le a \le 5$
      • $2 \le b \le 4$
      • $1 \le c \le 3$
      • $a + b + c = 7$

   2. List all possible triples

      Since the ranges are small, we enumerate systematically. Start with $a$ from 3 to 5, then find possible $b$ and $c$ that sum to 7.

      • If $a = 3$: then $b + c = 4$.

        Possible $(b,c)$:

        $b=2, c=2$ (since $c \ge 1$ and $b \le 4$)

        $b=3, c=1$

        $b=4, c=0$ — invalid because $c \ge 1$.

        So valid: (3,2,2) and (3,3,1).

      • If $a = 4$: then $b + c = 3$.

        Possible:

        $b=2, c=1$

        $b=3, c=0$ — invalid.

        So only (4,2,1).

      • If $a = 5$: then $b + c = 2$.

        Possible:

        $b=2, c=0$ — invalid.

        $b=1, c=1$ — but $b \ge 2$, so invalid.

        No valid triple for $a=5$.

      Thus the only valid triples are:

      $(3,2,2)$, $(3,3,1)$, $(4,2,1)$.

   3. Compute the number of ways for each triple

      For each triple, multiply the combinations from each part. …

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6. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.Bag A contains 4 white, 3 red and 2 black balls and bag B contains 2 white, 4 red and 3 black balls. If a bag is chosen at random and two balls are drawn at random from that bag, then the probability of drawing a white ball and a black ball is (A) $\dfrac{7}{18}$ (B) $\dfrac{2}{9}$ (C) $\dfrac{7}{36}$ (D) $\dfrac{1}{6}$

   ›Reveal solutionSolution

   The probability is found by conditioning on which bag is chosen (each with probability 1/2), then for each bag computing the probability of drawing one white and one black ball in two draws without replacement. The final result is 7/36, which corresponds to option (C).

   We have two bags with different compositions, and the bag is chosen at random. The event we care about — drawing one white and one black ball — depends on which bag we pick. So we use the law of total probability:

   $P(\text{white and black}) = P(\text{bag A}) \cdot P(\text{white and black} \mid A) + P(\text{bag B}) \cdot P(\text{white and black} \mid B)$

   Since each bag is equally likely, $P(\text{bag A}) = P(\text{bag B}) = \frac12$.

   Now we compute the conditional probabilities.

   1. Bag A has 4 white, 3 red, 2 black → total 9 balls. We want one white and one black in two draws without replacement. Number of ways to choose 2 balls from 9: $\binom{9}{2} = 36$. Number of favorable ways: choose 1 white from 4 and 1 black from 2 → $4 \times 2 = 8$. So

   $P(\text{white and black} \mid A) = \frac{8}{36} = \frac{2}{9}.$

   2. Bag B has 2 white, 4 red, 3 black → total 9 balls. Total ways: $\binom{9}{2} = 36$. Favorable ways: choose 1 white from 2 and 1 black from 3 → $2 \times 3 = 6$. So

   $P(\text{white and black} \mid B) = \frac{6}{36} = \frac{1}{6}.$

   3. Now combine using the law of total probability: $$$$

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7. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If all the letters of the word ‘ASSOCIATION’ are permuted in all possible ways to form all 11-letter words (with or without meaning), then among these words, the number of words in which the identical letters are always together but no two such groups of identical letters are together is (A) 1440 (B) 2304 (C) 576 (D) 144

   ›Reveal solutionSolution

   Grouping each repeated pair (AA, SS, OO, II) as a block and requiring no two blocks adjacent gives $3!\times 4! = 144$ words — option (D).

   The word ASSOCIATION has 11 letters: $A,S,O,I$ each appear twice, while $C,T,N$ appear once.

   "Identical letters are always together" means each repeated pair is tied into a single block: AA, SS, OO, II — four distinct blocks. The single letters $C,T,N$ (three distinct) serve as separators.

   "No two such groups of identical letters are together" means no two of the four blocks may be adjacent.

   Step 1 — arrange the separators. Place the 3 distinct single letters: $3! = 6$ ways. This creates 4 gaps (the two ends plus the two internal spaces):

   $\_\;C\;\_\;T\;\_\;N\;\_$ …

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8. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If two cards are drawn at a time from a well shuffled pack of 52 cards, then the probability of getting a result containing only one king and only one spade card is (A) $\dfrac{8}{221}$ (B) $\dfrac{49}{1326}$ (C) $\dfrac{12}{221}$ (D) $\dfrac{1}{26}$

   ›Reveal solutionSolution

   Count draws of two cards with exactly one king and exactly one spade in two disjoint cases ($36+36=72$), then divide by $\binom{52}{2}=1326$: $\dfrac{72}{1326}=\dfrac{12}{221}$ — option (C).

   Total outcomes. Drawing 2 of 52 cards (order irrelevant):

   $\binom{52}{2}=\frac{52\times51}{2}=1326.$

   Favourable outcomes. We need exactly one king and exactly one spade among the two cards. The king of spades belongs to both categories, so split into two disjoint cases.

   Case 1 — one card is the king of spades. It supplies the single king and the single spade, so the other card must be neither a king nor a spade. There are 3 other kings and 12 other spades, i.e. $15$ forbidden cards among the remaining $51$, leaving $51-15=36$ choices. …

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9. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.The number of positive integral solutions of $xyz = 60$ is (A) $_2C_{59}$ (B) $_2C^4 \times {}_2C^3 \times {}_2C^3$ (C) $_3C^4$ (D) $_1C^3 \times {}_0C^4 \times {}_4C^4$

   ›Reveal solutionSolution

   Distribute the prime exponents of $60=2^2\cdot3\cdot5$ among $x,y,z$: $\binom{4}{2}\binom{3}{2}\binom{3}{2}=54$ solutions. Option (B).

   Write $60 = 2^2\cdot 3^1\cdot 5^1$. For positive integers with $xyz=60$, split each prime's exponent among the three factors (stars and bars, nonnegative exponents):

   • exponent of $2$: $a_1+a_2+a_3=2 \Rightarrow \binom{2+2}{2}=\binom{4}{2}=6$
   • exponent of $3$: $b_1+b_2+b_3=1 \Rightarrow \binom{3}{2}=3$
   • exponent of $5$: $c_1+c_2+c_3=1 \Rightarrow \binom{3}{2}=3$ …

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10. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
    Q.The number of positive integral solutions of $xyz = 60$ is (A) ${}^4C_3$ (B) ${}^3C_1 \times {}^4C_0 \times {}^4C_4$ (C) ${}^4C_2 \times {}^3C_2 \times {}^3C_2$ (D) ${}^{59}C_2$

    ›Reveal solutionSolution

    The key idea is to count the number of ways to distribute the prime factors of 60 among three variables, which reduces to a stars-and-bars problem. The final answer is 54, which matches option (C).

    We need the number of positive integer solutions to $xyz = 60$. Since the variables are positive integers, each must be a divisor of 60. The natural approach is to factor 60 completely and then assign its prime factors to $x, y, z$.

    Why this works:

    The equation $xyz = 60$ means that the product of the three numbers equals 60. If we write 60 as a product of primes, then each prime factor must appear in exactly one of $x, y, z$ (or be split across them). Counting the number of ways to distribute the prime factors is equivalent to counting the number of ordered triples $(x, y, z)$ of positive integers whose product is 60.

    ---

    1. Factorize 60 completely:

       $60 = 2^2 \times 3^1 \times 5^1$.

       So we have: two 2’s, one 3, and one 5 to distribute among $x, y, z$.

    2. Treat each prime independently:

       Because the primes are distinct, the distribution of the 2’s does not affect the distribution of the 3’s or 5’s. The total number of ordered triples is the product of the number of ways to distribute each prime’s exponent.

    3. Distribute the two 2’s:

       We need to assign the exponent of 2 (which can be 0, 1, or 2) to each of $x, y, z$ such that the sum of the exponents equals 2. This is a stars-and-bars problem: number of nonnegative integer solutions to $a + b + c = 2$ is $\binom{2+3-1}{3-1} = \binom{4}{2} = 6$.

    4. Distribute the one 3:

       Similarly, for the exponent of 3, we need $d + e + f = 1$ (nonnegative integers). Number of solutions: $\binom{1+3-1}{3-1} = \binom{3}{2} = 3$.

    5. Distribute the one 5:

       Same as for 3: $g + h + i = 1$ gives $\binom{3}{2} = 3$ ways.

    6. Multiply the counts:

       Total ordered triples = $6 \times 3 \times 3 = 54$.

    Now check the options: …

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11. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.Three similar urns A, B, C contain 2 red and 3 white balls; 3 red and 2 white balls; 1 red and 4 white balls respectively. If a ball selected at random from one of the urns is found to be red, then the probability that it is drawn from urn C is (A) $\dfrac{1}{6}$ (B) $\dfrac{1}{3}$ (C) $\dfrac{1}{2}$ (D) $\dfrac{2}{9}$

    ›Reveal solutionSolution

    This is a classic Bayes’ theorem problem: we update the prior probability of picking each urn (each 1/3) with the likelihood of drawing a red ball from that urn. The posterior probability that the red ball came from urn C is 1/6, so the correct option is (A).

    We are told there are three urns, each with a different mix of red and white balls.

    • Urn A: 2 red, 3 white → total 5 balls.
    • Urn B: 3 red, 2 white → total 5 balls.
    • Urn C: 1 red, 4 white → total 5 balls.

    A ball is selected at random from one of the urns (each urn equally likely), and it turns out to be red. We want the probability that it came from urn C.

    Concept & Intuition

    This is a textbook case for Bayes’ theorem. We have three “causes” (the urn chosen) and one “effect” (a red ball). We know the probability of the effect given each cause (the fraction of red balls in that urn). We also know the prior probability of each cause (each urn is equally likely). Bayes’ theorem lets us reverse the conditional: given the effect occurred, what’s the probability it came from a particular cause?

    The key is to compute the total probability of drawing a red ball (the denominator) and then the portion contributed by urn C (the numerator).

    ---

    Step-by-step solution

    1. Define events

       Let $R$ be the event “the drawn ball is red.”

       Let $A, B, C$ be the events that the ball came from urn A, B, or C respectively.

    2. Write down the prior probabilities

       Since the urn is chosen at random,

    $P(A) = P(B) = P(C) = \frac{1}{3}.$

    3. Write down the likelihoods (probability of drawing a red ball from each urn)

       • From urn A: 2 red out of 5 → $P(R \mid A) = \frac{2}{5}$.
       • From urn B: 3 red out of 5 → $P(R \mid B) = \frac{3}{5}$.
       • From urn C: 1 red out of 5 → $P(R \mid C) = \frac{1}{5}$.

    4. Compute the total probability of drawing a red ball (the denominator in Bayes’ theorem)

       By the law of total probability:

    $P(R) = P(A)P(R\mid A) + P(B)P(R\mid B) + P(C)P(R\mid C)$

    Substitute:

    $$P(R) = \frac{1}{3}\cdot\frac{2}{5} + \frac{1}{3}\cdot\frac{3}{5} + \frac{1}{3}\cdot\frac{1}{5} = \frac{1}{3}\left(\frac{2+3+1}{5}\right) = \frac{1}{3}\cdot\frac{6}{5} …$$

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12. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.A fair coin is tossed a fixed number of times. If the probability of getting 5 heads is equal to the probability of getting 4 heads, then the probability of getting 6 heads is (A) $\dfrac{7}{64}$ (B) $\dfrac{9}{32}$ (C) $\dfrac{21}{128}$ (D) $\dfrac{35}{256}$

    ›Reveal solutionSolution

    The key idea is that for a binomial distribution with equal head/tail probability, equal probabilities for 4 and 5 heads implies the number of trials is $n = 9$. Then the probability of 6 heads is $\binom{9}{6}(1/2)^9 = 84/512 = 21/128$, which corresponds to option (C).

    We are told a fair coin is tossed a fixed number $n$ times. The probability of getting exactly $k$ heads in $n$ tosses is given by the binomial formula:

    $P(k) = \binom{n}{k} \left(\frac12\right)^n$

    since the coin is fair (probability of heads = $1/2$).

    The condition: $P(5) = P(4)$. That means:

    $\binom{n}{5} \left(\frac12\right)^n = \binom{n}{4} \left(\frac12\right)^n$

    Cancelling the common factor $(1/2)^n$ (which is never zero), we get:

    $\binom{n}{5} = \binom{n}{4}$

    Now, recall the symmetry property of binomial coefficients: $\binom{n}{k} = \binom{n}{n-k}$. But here we have equality between consecutive numbers, not symmetric ones. So we use the explicit formula:

    $\binom{n}{5} = \frac{n!}{5!(n-5)!}, \quad \binom{n}{4} = \frac{n!}{4!(n-4)!}$

    Setting them equal:

    $\frac{n!}{5!(n-5)!} = \frac{n!}{4!(n-4)!}$

    Cancel $n!$ and multiply both sides by $5! (n-4)!$:

    $(n-4)! = 5 \cdot (n-5)!$

    Since $(n-4)! = (n-4)(n-5)!$, we have:

    $(n-4)(n-5)! = 5 (n-5)!$

    Cancel $(n-5)!$ (nonzero for $n \ge 5$):

    $n - 4 = 5 \quad \Rightarrow \quad n = 9$

    So the coin is tossed 9 times.

    Now we want $P(6)$:

    $P(6) = \binom{9}{6} \left(\frac12\right)^9$ …

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