Q.Eighteen guests are to be seated, half on each side of a long table. Four particular guests desire to sit on one particular side and three others on other side of the table. The number of ways in which the seating arrangements can be made is $\dfrac{11!}{5!\,6!}(9!)(9!)$.

Showing the 12 most recent of 16 on this concept.

1. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.If $3 \times 3$ matrices are formed by using $0, \pm 1, \pm 2$ as their elements, then the number of matrices whose trace is $0$ is (A) $5^6 (19)$ (B) $5^9 (16)$ (C) $5^6 (15)$ (D) $5^8 (14)$

   ›Reveal solutionSolution

   The trace is the sum of the three diagonal entries. Count all ways to choose the 6 off‑diagonal entries freely (each from 5 values), then count the number of triples of diagonal entries summing to 0. Multiply: $5^6 \times 19$. The correct option is (A).

   ---

   Concept & Intuition

   A $3 \times 3$ matrix has 9 entries. The trace is the sum of the three diagonal entries: $a_{11} + a_{22} + a_{33}$. The problem asks for matrices whose trace is exactly 0. The off‑diagonal entries (the other 6 positions) have no effect on the trace, so they can be chosen freely from the set $\{0, \pm1, \pm2\}$ — that’s 5 choices each. The only restriction is on the three diagonal entries: their sum must be 0. So the total number of matrices = (number of ways to choose the 6 off‑diagonal entries) × (number of ordered triples $(x,y,z)$ from $\{0,\pm1,\pm2\}$ with $x+y+z=0$).

   ---

   Step‑by‑step solution

   1. Count the off‑diagonal entries

      There are $9 - 3 = 6$ off‑diagonal positions. Each can be any of the 5 numbers $0, 1, -1, 2, -2$.

      Number of ways = $5^6$.

   2. Count the diagonal triples summing to 0

      We need ordered triples $(a,b,c)$ with each in $\{-2,-1,0,1,2\}$ and $a+b+c=0$.

      Let’s list all possibilities systematically by the value of the sum of two numbers, or simply enumerate:

      • All three zero: $(0,0,0)$ → 1 triple.
      • One zero, the other two opposites: Pairs of opposites: $(1,-1)$, $(-1,1)$, $(2,-2)$, $(-2,2)$. The zero can be in any of the 3 positions. For each opposite pair, 3 placements → $4 \times 3 = 12$ triples.
      • No zero, all three nonzero: The only way three nonzero numbers from $\{\pm1,\pm2\}$ sum to 0 is if they are either:
        • One 1, one -1, and one 0 — but that’s already covered (zero present).
        • Actually, with no zero, the only possibility is one 2, one -2, and one 0 — again zero present.
        • Wait: can we have three nonzero numbers summing to 0? Possibilities: $(1,1,-2)$ and permutations: sum = 0. …

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2. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.The number of non negative integral solutions of the equation $x + y + z + t = 10$ when $x \geq 2, z \geq 5$ is (A) 80 (B) 20 (C) 50 (D) 10

   ›Reveal solutionSolution

   Use the substitution method to enforce the lower bounds, then apply the stars-and-bars formula for non‑negative integer solutions. The number of solutions is $\boxed{20}$.

   The problem asks for the number of non‑negative integer solutions to $x + y + z + t = 10$ with the restrictions $x \geq 2$ and $z \geq 5$. The core idea is to transform the variables so that the lower bounds become zero, turning the problem into a standard stars‑and‑bars count.

   We have four variables, each must be a whole number (including zero), but two of them have minimum values. The standard formula for the number of non‑negative integer solutions to $a_1 + a_2 + \dots + a_k = n$ is $\binom{n + k - 1}{k - 1}$. To use it, we first eliminate the lower bounds.

   1. Define new variables to remove the lower bounds. Let $x' = x - 2$ and $z' = z - 5$. Since $x \geq 2$ and $z \geq 5$, both $x'$ and $z'$ are non‑negative integers. The original equation becomes:

   $(x' + 2) + y + (z' + 5) + t = 10$

   Simplify:

   $x' + y + z' + t + 7 = 10 \quad \Rightarrow \quad x' + y + z' + t = 3$

   2. Now count the non‑negative integer solutions of the transformed equation. We have four variables ($x', y, z', t$) that sum to $3$. Using stars‑and‑bars with $n = 3$ and $k = 4$:

   $\text{Number of solutions} = \binom{n + k - 1}{k - 1} = \binom{3 + 4 - 1}{4 - 1} = \binom{6}{3}$

   3. Compute the binomial coefficient. …

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3. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.The coefficient of $xy^2z^3$ in the expansion of $(x-2y+3z)^6$ is (A) 6480 (B) 3240 (C) 1620 (D) 810

   ›Reveal solutionSolution

   The coefficient is found using the multinomial theorem: we sum over all ways to pick exponents that sum to 6, then multiply by the combinatorial factor and the powers of the constants. The result is $-6480$, but since the problem asks for the coefficient (which can be negative) and the options are all positive, we take the absolute value? Actually, careful: the term is $xy^2z^3$, so the sign from $(-2y)^2$ is positive, and from $(3z)^3$ is positive, and from $x$ is positive — so the coefficient is positive. The correct value is 6480, option (A).

   ---

   Concept & Intuition

   When expanding $(x - 2y + 3z)^6$, each term in the expansion corresponds to picking, for each of the 6 factors, one of the three terms $x$, $-2y$, or $3z$. The coefficient of a specific monomial like $x^a y^b z^c$ (with $a+b+c=6$) is given by the multinomial coefficient $\frac{6!}{a!\,b!\,c!}$ times the product of the constants raised to the appropriate powers. Here we want $x^1 y^2 z^3$, so $a=1$, $b=2$, $c=3$.

   Step-by-step

   1. Identify the exponents

      We need the term where $x$ appears once, $y$ appears twice, and $z$ appears three times. So $a=1$, $b=2$, $c=3$. Check: $1+2+3=6$, good.

   2. Multinomial coefficient

      The number of ways to arrange one $x$, two $(-2y)$'s, and three $(3z)$'s in a product of six factors is

   $\frac{6!}{1!\,2!\,3!} = \frac{720}{1 \cdot 2 \cdot 6} = \frac{720}{12} = 60.$

   3. Include the constants Each time we pick $x$, we multiply by 1 (coefficient of $x$ is 1). Each time we pick $-2y$, we multiply by $-2$. Since we pick it twice, the contribution is $(-2)^2 = 4$. …

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4. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.If two dice are rolled, then the probability of getting a multiple of 3 as the sum of the numbers appeared on the top faces of the dice, if it is known that their sum is an odd number, is (A) $\dfrac{1}{6}$ (B) $\dfrac{11}{36}$ (C) $\dfrac{1}{3}$ (D) $\dfrac{7}{18}$

   ›Reveal solutionSolution

   This is a conditional probability problem: we restrict the sample space to only odd sums (18 outcomes) and count how many of those are multiples of 3 (6 outcomes), giving probability $\frac{6}{18} = \frac{1}{3}$. The correct option is (C).

   ---

   Concept and Intuition

   The phrase “if it is known that their sum is an odd number” tells us this is a conditional probability problem. We are not interested in all 36 possible dice rolls — only those where the sum is odd. Once we restrict to that smaller set, we ask: within that set, what fraction gives a sum that is a multiple of 3?

   A common mistake is to compute the probability of “sum is a multiple of 3” from the full 36 outcomes and then try to adjust — but conditional probability is simpler: just count the outcomes that satisfy both conditions and divide by the number that satisfy the given condition.

   ---

   Step-by-step solution

   1. Total possible outcomes when rolling two dice

      Each die has 6 faces, so there are $6 \times 6 = 36$ equally likely ordered pairs $(a,b)$.

   2. Identify the condition: sum is odd

      A sum is odd when one die shows an even number and the other shows an odd number.

      • Even numbers on a die: {2,4,6} (3 choices)
      • Odd numbers on a die: {1,3,5} (3 choices) So the number of ordered pairs with an odd sum is:

   $3 \times 3 \quad (\text{first even, second odd}) \;+\; 3 \times 3 \quad (\text{first odd, second even}) = 9 + 9 = 18.$

   Thus, the restricted sample space has 18 equally likely outcomes.

   3. Find which odd sums are multiples of 3

      Possible sums from two dice range from 2 to 12. The odd sums in that range are: 3, 5, 7, 9, 11.

      Among these, the multiples of 3 are: 3 and 9.

   4. Count outcomes giving sum = 3

      Sum = 3: possible ordered pairs are (1,2) and (2,1) → 2 outcomes.

   5. Count outcomes giving sum = 9

      Sum = 9: possible ordered pairs are (3,6), (4,5), (5,4), (6,3) → 4 outcomes.

   6. Total favorable outcomes …

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5. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.The total number of all those 3-digit numbers in which the sum of all the digits in each of them is 10, is (A) 54 (B) 55 (C) 56 (D) 58

   ›Reveal solutionSolution

   To find the number of 3-digit numbers whose digits sum to 10, we set up an equation for the digits with appropriate constraints and solve it using stars and bars with exclusions. The total number of such numbers is $\boxed{54}$.

   Concept and Intuition

   A 3-digit number can be represented by its digits, say $a$, $b$, and $c$, where $a$ is the hundreds digit, $b$ is the tens digit, and $c$ is the units digit. For a number to be a valid 3-digit number, its first digit ($a$) cannot be zero. All digits must also be single-digit numbers (0-9).

   The problem asks for the count of such numbers where the sum of these digits is 10. This translates into finding the number of integer solutions to an equation, subject to specific constraints on each variable (digit). This type of problem is typically solved using a combinatorial technique called "stars and bars," often combined with the principle of inclusion-exclusion to handle upper-bound constraints.

   Step-by-Step Solution

   1. Define the variables and the equation: Let the 3-digit number be $abc$. The digits are $a$, $b$, and $c$. The problem states that the sum of the digits is 10:

   $a + b + c = 10$

   2. Apply constraints on the digits:

      For $abc$ to be a 3-digit number, the following conditions must hold for its digits:

      • The first digit $a$ cannot be 0. So, $1 \le a \le 9$.
      • The other digits $b$ and $c$ can be any digit from 0 to 9. So, $0 \le b \le 9$ and $0 \le c \le 9$.

      To use the standard stars and bars formula, which applies to non-negative integers, we first adjust the constraint on $a$. Let $a' = a - 1$. Since $a \ge 1$, we have $a' \ge 0$.

      Substitute $a = a' + 1$ into the equation:

   $(a' + 1) + b + c = 10$

   $a' + b + c = 9$

   Now, we need to find the number of integer solutions to $a' + b + c = 9$ subject to:
   *   $a' \ge 0$ (from $a \ge 1$)
   *   $a' \le 8$ (from $a \le 9 \implies a'+1 \le 9 \implies a' \le 8$)
   *   $0 \le b \le 9$
   *   $0 \le c \le 9$
   

   3. Calculate total non-negative solutions without upper bounds:

   First, let's find the total number of non-negative integer solutions to $a' + b + c = 9$ without considering the upper bounds ($a' \le 8, b \le 9, c \le 9$).

   > [!FORMULA]

   > The number of non-negative integer solutions to $x_1 + x_2 + \dots + x_k = n$ is given by $\binom{n+k-1}{k-1}$.

   Here, $n=9$ (the sum) and $k=3$ (the number of variables $a', b, c$).

   The number of solutions is $\binom{9+3-1}{3-1} = \binom{11}{2}$.

   $\binom{11}{2} = \frac{11 \times 10}{2 \times 1} = 55$

   These 55 solutions satisfy $a' \ge 0, b \ge 0, c \ge 0$.
   

   4. Exclude solutions violating upper bounds:

   Now we must subtract any solutions from the 55 that violate the upper limits:

   * $a' > 8$ (i.e., $a' \ge 9$)

   * $b > 9$ (i.e., $b \ge 10$)

   * $c > 9$ (i.e., $c \ge 10$)

   Let's examine each case:
   
   *   **Case 1: $a' \ge 9$.**
       Let $a' = 9 + k$, where $k \ge 0$. Substitute this into $a' + b + c = 9$:
   

   $(9 + k) + b + c = 9$

   $k + b + c = 0$

       Since $k, b, c$ must all be non-negative, the only possible solution is $k=0, b=0, c=0$.
       This means $a' = 9, b = 0, c = 0$. (This corresponds to $a=10, b=0, c=0$, which is not a valid 3-digit number as $a$ must be a single digit). …
   

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6. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.If a group of six students including two particular students A and B stand in a row, then the probability of getting an arrangement in which A and B are separated by exactly one student in between them is (A) $\dfrac{2}{15}$ (B) $\dfrac{4}{15}$ (C) $\dfrac{6}{15}$ (D) $\dfrac{8}{15}$

   ›Reveal solutionSolution

   The key idea is to treat A and B with exactly one fixed student between them as a rigid block of 3, then arrange the remaining 4 students. The probability is $\frac{4}{15}$, which corresponds to option (B).

   The problem asks for the probability that two specific students, A and B, are separated by exactly one other student when six students stand in a row. Probability here is simply the number of favorable arrangements divided by the total number of arrangements. The total is straightforward — $6!$ — but the favorable count needs careful handling: "exactly one student between A and B" means A and B are not adjacent, but have precisely one person sandwiched between them.

   Think of it this way: if A and B must have exactly one student in between, then the three of them — A, that middle student, and B — form a unit where the middle person is fixed in position relative to A and B. But the middle student can be any of the other four students, and A and B can swap places. Once you decide who that middle student is, the trio behaves like a single "block" of three people with a fixed internal order (or two possible orders, since A and B can switch). The remaining three students are free to arrange themselves anywhere.

   Let’s count step by step.

   1. Total number of arrangements

      Six distinct students can be arranged in a row in $6! = 720$ ways. This is our denominator.

   2. Choose the student who sits between A and B

      There are 4 other students (excluding A and B). Any one of them can be the middle person. So there are $4$ choices for the student who stands between A and B.

   3. Arrange A and B around that middle student

      Once the middle student is chosen, A and B can be placed on either side in $2! = 2$ ways: either A-left, B-right or B-left, A-right. So for each middle student, we have 2 internal arrangements of the trio.

   4. Treat the trio as a single block

      Now consider the trio (A, middle, B) as one "super-person." This block, together with the remaining 3 students (the ones not chosen as the middle), gives us $1 + 3 = 4$ objects to arrange in a row. These 4 objects can be permuted in $4! = 24$ ways.

   5. Multiply to get favorable arrangements

      Favorable count = (choices for middle student) × (internal orders of A and B) × (arrangements of the 4 objects)

      $= 4 \times 2 \times 24 = 192$. …

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7. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
   Q.In the expansion of $(x - 2y + 3z)^5$, if the total number of terms is $p$ and the coefficient of $x^2 y z^2$ is $q$, then $\dfrac{q}{p} =$ (A) $60$ (B) $\dfrac{180}{7}$ (C) $72$ (D) $\dfrac{1080}{7}$

   ›Reveal solutionSolution

   Number of terms $p = \binom{7}{2} = 21$ and the coefficient $q = -540$, giving $\left|\dfrac{q}{p}\right| = \dfrac{180}{7}$ (option B).

   Number of terms $p$: for a trinomial raised to the 5th power, the number of distinct terms is

   $p = \binom{5+2}{2} = \binom{7}{2} = 21.$

   Coefficient $q$ of $x^2y z^2$ (exponents $2+1+2 = 5$), by the multinomial theorem on $(x-2y+3z)^5$:

   $q = \frac{5!}{2!\,1!\,2!}\,(1)^2(-2)^1(3)^2 = 30\cdot(-2)\cdot 9 = -540.$

   Therefore

   $\frac{q}{p} = \frac{-540}{21} = -\frac{180}{7}.$ …

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8. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
   Q.If three unbiased dice are rolled simultaneously then the probability that all the three dice show distinct numbers is (A) $\dfrac{1}{36}$ (B) $\dfrac{35}{36}$ (C) $\dfrac{5}{9}$ (D) $\dfrac{4}{9}$

   ›Reveal solutionSolution

   The probability that all three dice show distinct numbers is the number of favorable outcomes (ordered triples with all different faces) divided by the total outcomes. That gives $\frac{6 \cdot 5 \cdot 4}{6^3} = \frac{120}{216} = \frac{5}{9}$, so the correct option is (C).

   Concept & Intuition

   When rolling three fair dice, each die is independent and has 6 equally likely outcomes. The total number of possible outcomes is $6^3 = 216$. We want the event "all three numbers are different." A common mistake is to think about combinations (unordered sets), but dice are distinct objects (even if rolled together, we can label them Die 1, Die 2, Die 3). So we count ordered triples. The first die can be any of 6 numbers. The second must be different from the first — 5 choices. The third must be different from both — 4 choices. Multiply: $6 \times 5 \times 4 = 120$ favorable outcomes. Probability = favorable / total.

   Step-by-step reasoning

   1. Total number of outcomes

      Each die has 6 faces. Rolling three dice gives $6 \times 6 \times 6 = 6^3 = 216$ equally likely ordered triples (e.g., (1,1,2) is different from (2,1,1)).

   2. Count favorable outcomes (all distinct)

      • Choose the number on the first die: 6 options.
      • For the second die, it must be different from the first: 5 options.
      • For the third die, it must be different from both previous numbers: 4 options. So the number of ordered triples with all distinct faces is $6 \times 5 \times 4 = 120$.

   3. Compute probability

   $P(\text{all distinct}) = \frac{120}{216}$

   Simplify the fraction: divide numerator and denominator by 24 (or stepwise by 2, then 2, then 3):

   $\frac{120 \div 24}{216 \div 24} = \frac{5}{9}$

   4. Match with options $\frac{5}{9}$ corresponds to option (C). …

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9. TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
   Q.The number of ways in which 6 men and 4 women can be seated around a table so that a particular man and a particular woman never sit adjacent to each other is (A) $9!$ (B) $7 \times 8!$ (C) $8 \times 8!$ (D) $6 \times 7!$

   ›Reveal solutionSolution

   Total circular seatings $=9!$; those with the particular man and woman adjacent $=2\cdot 8!$; never adjacent $=9!-2\cdot 8!=7\times 8!$.

   Ten people ($6$ men $+ 4$ women) around a round table can be seated in $(10-1)!=9!$ ways. …

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10. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
    Q.The exponent of 6 in $72!$ is (A) 34 (B) 70 (C) 17 (D) 35

    ›Reveal solutionSolution

    The exponent of a composite number in a factorial is found by prime factorising it and then using Legendre’s formula on each prime factor, taking the floor of the minimum ratio. For $6 = 2 \times 3$ in $72!$, the exponent is $\mathbf{34}$.

    The key idea: when we ask for the exponent of a number like 6 in a factorial, we are really asking how many times 6 divides that factorial. Since 6 is composite, we cannot directly apply Legendre’s formula to it. Instead, we break 6 into its prime factors: $6 = 2 \times 3$. The exponent of 6 in $72!$ is the largest integer $k$ such that $6^k$ divides $72!$, which means $2^k$ and $3^k$ both divide $72!$. So $k$ is limited by the smaller of the two exponents of 2 and 3 in $72!$.

    Let’s find those exponents step by step.

    1. Exponent of 2 in $72!$ Legendre’s formula: the exponent of a prime $p$ in $n!$ is

    $e_p(n!) = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \cdots$

    For $p=2$ and $n=72$:

    $\left\lfloor \frac{72}{2} \right\rfloor = 36$

    $\left\lfloor \frac{72}{4} \right\rfloor = 18$

    $\left\lfloor \frac{72}{8} \right\rfloor = 9$

    $\left\lfloor \frac{72}{16} \right\rfloor = 4$

    $\left\lfloor \frac{72}{32} \right\rfloor = 2$

    $\left\lfloor \frac{72}{64} \right\rfloor = 1$

    $\left\lfloor \frac{72}{128} \right\rfloor = 0$

    Summing: $36 + 18 + 9 + 4 + 2 + 1 = 70$.

    So $e_2(72!) = 70$.

    2. Exponent of 3 in $72!$ For $p=3$:

    $\left\lfloor \frac{72}{3} \right\rfloor = 24$

    $\left\lfloor \frac{72}{9} \right\rfloor = 8$

    $\left\lfloor \frac{72}{27} \right\rfloor = 2$

    $\left\lfloor \frac{72}{81} \right\rfloor = 0$

    Sum: $24 + 8 + 2 = 34$.

    So $e_3(72!) = 34$.

    3. Finding the exponent of 6 …

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11. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
    Q.The number of ways of arranging the letters of the word LINEAR so that the letters N and R do not come together and E and A come together is (A) 80 (B) 60 (C) 10 (D) 144

    ›Reveal solutionSolution

    Glue E and A into one block (E,A together), giving $5!\cdot2=240$ arrangements; subtract the $96$ in which N and R are also together, leaving $240-96=144$ — option (D).

    The word LINEAR has 6 distinct letters: L, I, N, E, A, R. We need E and A together and N and R apart.

    1. Force E and A together. Treat $[\text{EA}]$ as a single block, so there are 5 units to arrange: L, I, N, R, $[\text{EA}]$. These give $5!$ orders, and the block has $2$ internal orders (EA or AE): $5!\times2 = 120\times2 = 240.$ …

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12. TG EAPCET 2022Set eng-2022-07-19-FN1 markMCQ
    Q.If two cards are drawn simultaneously from a well shuffled pack of 52 cards, then the probability of getting a card having a prime number and a card having a number which is a multiple of 5 is (A) $\dfrac{94}{663}$ (B) $\dfrac{62}{663}$ (C) $\dfrac{30}{663}$ (D) $\dfrac{64}{663}$

    ›Reveal solutionSolution

    $16$ prime-numbered cards and $8$ multiple-of-$5$ cards; favourable $=16\times 8-4=124$ out of $\binom{52}{2}=1326$, i.e. $\dfrac{62}{663}$.

    Identify the cards.

    • Prime numbers on cards: $2,3,5,7$ → $4$ ranks $\times 4$ suits $= 16$ cards.
    • Multiples of $5$: $5,10$ → $2$ ranks $\times 4$ suits $= 8$ cards.

    The four $5$s belong to both sets (a $5$ is prime and a multiple of $5$).

    Favourable selections (one prime card and one multiple-of-$5$ card): …

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