Q.Evaluate $\dfrac{n!}{(n-r)!}$, when

Showing the 12 most recent of 19 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If $n$ is an integer between $0$ and $31$, then the minimum value of $n!(31-n)!$ is (A) $10! \, 21!$ (B) $15! \, 16!$ (C) $30!$ (D) $31!$

   ›Reveal solutionSolution

   $n!(31-n)!$ is minimized exactly where $\binom{31}{n}$ is maximized — the middle of the range $0\le n\le31$ — giving $15!\,16!$.

   Concept and Intuition

   Since $\binom{31}{n}=\dfrac{31!}{n!(31-n)!}$ and $31!$ is fixed, minimizing $n!(31-n)!$ is the same as maximizing $\binom{31}{n}$. Binomial coefficients $\binom{N}{n}$ are known to be largest at the centre of the range $n=0,\dots,N$ and to decrease symmetrically toward the ends — so the minimum of $n!(31-n)!$ occurs at the centre, not at the extremes.

   Step-by-Step Solution

   1. We want to minimize $f(n)=n!(31-n)!$ for integer $n$ in $[0,31]$.
   2. Note $f(n) = \dfrac{31!}{\binom{31}{n}}$, so minimizing $f(n)$ is equivalent to maximizing $\binom{31}{n}$.
   3. For $\binom{N}{n}$, the maximum occurs at $n=\lfloor N/2\rfloor$ (and also at $n=\lceil N/2\rceil$ if $N$ is odd, by the symmetry $\binom{N}{n}=\binom{N}{N-n}$).
   4. Here $N=31$ is odd, so the two largest, equal binomial coefficients occur at $n=15$ and $n=16$: $\binom{31}{15}=\binom{31}{16}$.
   5. So the minimum of $f(n)=n!(31-n)!$ occurs at $n=15$ (or equivalently $n=16$): $f(15)=15!\times16!$. …

   🔒 Start your 14-day free trial to see the full solution →
2. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.The exponent of 6 in 72! is (A) $14$ (B) $16$ (C) $34$ (D) $70$

   ›Reveal solutionSolution

   The exponent of a composite number $6=2\cdot3$ in $n!$ is limited by whichever prime factor is scarcer — here, the exponent of 3, since 3's exponent (34) is much smaller than 2's (70).

   Concept and Intuition

   When a number is not prime, its exponent (the highest power dividing $n!$) is governed by the least available exponent among its prime factors, since you can only form as many complete '6's as you can pair a 2 with a 3, and 3s are always scarcer than 2s in $n!$ for $n\ge 2$. Legendre's formula gives the exponent of a prime $p$ in $n!$ as $\sum_{k\ge1}\left\lfloor \dfrac{n}{p^k}\right\rfloor$.

   Step-by-Step Solution

   1. Exponent of 2 in $72!$: $\left\lfloor\frac{72}{2}\right\rfloor+\left\lfloor\frac{72}{4}\right\rfloor+\left\lfloor\frac{72}{8}\right\rfloor+\left\lfloor\frac{72}{16}\right\rfloor+\left\lfloor\frac{72}{32}\right\rfloor+\left\lfloor\frac{72}{64}\right\rfloor = 36+18+9+4+2+1=70$. …

   🔒 Start your 14-day free trial to see the full solution →
3. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.If ${}^{n}P_4 = 5040$ and ${}^{15}P_r = 2730$, then ${}^{n}P_r =$ (A) 120 (B) 720 (C) 1680 (D) 840

   ›Reveal solutionSolution

   Match $^nP_4=5040$ to $n=10$ and $^{15}P_r=2730$ to $r=3$ by recognizing the products of consecutive integers, then compute $^{10}P_3=720$.

   Concept and Intuition

   $^nP_k$ is just the product of $k$ consecutive integers counting down from $n$. Rather than solving factorial equations algebraically, it's much faster to recognize these products directly by testing nearby integers, since permutation values grow fast and are easy to bracket.

   Step-by-Step Solution

   1. $^nP_4 = n(n-1)(n-2)(n-3) = 5040$. Testing $n=10$: $10\times9\times8\times7 = 5040$. ✓ So $n=10$.
   2. $^{15}P_r = 15\times14\times\cdots\times(16-r) = 2730$. Testing $r=3$: $15\times14\times13 = 2730$. ✓ So $r=3$.
   3. Now compute $^{n}P_r = {}^{10}P_3 = 10\times9\times8 = 720$. …

   🔒 Start your 14-day free trial to see the full solution →
4. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If $^{27}P_{r+7} = 7722\ ^{25}P_{(r+4)}$, then $r =$ (A) 9 (B) 12 (C) 11 (D) 10

   ›Reveal solutionSolution

   Write both permutation expressions using $^nP_k=\dfrac{n!}{(n-k)!}$, simplify the factorial ratio, and solve the resulting linear equation for $r$. Answer: $r=10$.

   Concept and Intuition

   Permutation expressions with a variable in both the upper and lower index often simplify beautifully once you write everything as factorial ratios — most of the factorial cancels, leaving a simple linear (or low-degree) equation.

   Step-by-Step Solution

   1. $^{27}P_{r+7}=\dfrac{27!}{(27-(r+7))!}=\dfrac{27!}{(20-r)!}$.
   2. $^{25}P_{r+4}=\dfrac{25!}{(25-(r+4))!}=\dfrac{25!}{(21-r)!}$.
   3. The given equation: $\dfrac{27!}{(20-r)!}=7722\cdot\dfrac{25!}{(21-r)!}$.
   4. Write $27!=27\times26\times25!$, so $\dfrac{27!}{25!}=702$. The equation becomes $702\cdot\dfrac{25!}{(20-r)!}=7722\cdot\dfrac{25!}{(21-r)!}$, and $25!$ cancels.
   5. So $\dfrac{702}{(20-r)!}=\dfrac{7722}{(21-r)!}$, i.e. $\dfrac{(21-r)!}{(20-r)!}=\dfrac{7722}{702}=11$. …

   🔒 Start your 14-day free trial to see the full solution →
5. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.The number of divisors of $7!$ is (A) 72 (B) 24 (C) 64 (D) 60

   ›Reveal solutionSolution

   Prime-factorizing $7!=2^4\cdot3^2\cdot5\cdot7$ and applying the divisor-count formula $(e_1+1)(e_2+1)\cdots$ gives $60$ divisors.

   Concept and Intuition

   If $n=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$, the total number of positive divisors of $n$ is $(e_1+1)(e_2+1)\cdots(e_k+1)$, since each divisor is formed by independently choosing an exponent from $0$ to $e_i$ for each prime.

   Step-by-Step Solution

   1. $7! = 1\times2\times3\times4\times5\times6\times7 = 5040$.
   2. Prime factorize by counting powers of each prime $\le7$ among $1,\dots,7$:
      • Power of $2$: from $2,4=2^2,6=2\times3$ → $2^{1+2+1}=2^4$.
      • Power of $3$: from $3, 6=2\times3$ → $3^{1+1}=3^2$.
      • Power of $5$: from $5$ → $5^1$.
      • Power of $7$: from $7$ → $7^1$. So $7! = 2^4\cdot3^2\cdot5^1\cdot7^1$ (check: $16\times9\times5\times7 = 144\times35=5040$ ✓). …

   🔒 Start your 14-day free trial to see the full solution →
6. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.$\displaystyle\sum_{r=1}^{15} r^2\left(\dfrac{{}^{15}C_r}{{}^{15}C_{r-1}}\right)=$ (A) $560$ (B) $680$ (C) $840$ (D) $1020$

   ›Reveal solutionSolution

   Simplify the binomial-coefficient ratio first, turning the sum into a simple polynomial sum; the total is $680$.

   Concept and Intuition

   The ratio $\dfrac{{}^nC_r}{{}^nC_{r-1}}=\dfrac{n-r+1}{r}$ is a standard simplification that converts an awkward combinatorial sum into ordinary arithmetic series.

   Step-by-Step Solution

   1. With $n=15$: $\dfrac{{}^{15}C_r}{{}^{15}C_{r-1}}=\dfrac{15-r+1}{r}=\dfrac{16-r}{r}$.
   2. So each term $r^2\cdot\dfrac{16-r}{r}=r(16-r)=16r-r^2$.
   3. Sum from $r=1$ to $15$: $\displaystyle\sum 16r-\sum r^2=16\cdot\frac{15\cdot16}{2}-\frac{15\cdot16\cdot31}{6}$. …

   🔒 Start your 14-day free trial to see the full solution →
7. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.The greatest integer r such that $30^r$ divides $30!$ is (A) $8$ (B) $7$ (C) $6$ (D) $5$

   ›Reveal solutionSolution

   $30^r=2^r3^r5^r$ divides $30!$ only as far as the scarcest prime factor allows — and $5$ appears only $7$ times in $30!$, so $r=7$.

   Concept and Intuition

   Legendre's formula gives the exact power of a prime $p$ dividing $n!$: $\sum_{k\ge1}\lfloor n/p^k\rfloor$. Since $30=2\times3\times5$, $30^r$ needs $r$ copies of each of $2,3,5$ simultaneously; the answer is the minimum of the three individual prime-power counts in $30!$.

   Step-by-Step Solution

   1. Power of $5$ in $30!$: $\lfloor30/5\rfloor+\lfloor30/25\rfloor=6+1=7$.
   2. Power of $3$ in $30!$: $\lfloor30/3\rfloor+\lfloor30/9\rfloor+\lfloor30/27\rfloor=10+3+1=14$.
   3. Power of $2$ in $30!$: $\lfloor30/2\rfloor+\lfloor30/4\rfloor+\lfloor30/8\rfloor+\lfloor30/16\rfloor=15+7+3+1=26$. …

   🔒 Start your 14-day free trial to see the full solution →
8. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.The number of arrangements of the letters of the word ARRANGEMENT in which two Es' do not occur adjacently is (A) $\dfrac{9}{8}(10)!$ (B) $\dfrac{9}{4}(10)!$ (C) $\dfrac{9}{16}(10)!$ (D) $\dfrac{9}{32}(10)!$

   ›Reveal solutionSolution

   Total arrangements minus the "E's glued together" arrangements gives the count with the two E's never adjacent: $\frac{9}{16}\cdot10!$.

   Concept and Intuition

   The standard technique for "certain letters must NOT be adjacent" is: (Total arrangements) $-$ (arrangements where they ARE adjacent, found by gluing them into a single block). Both counts must account for the word's repeated letters correctly.

   Step-by-Step Solution

   1. ARRANGEMENT = A,R,R,A,N,G,E,M,E,N,T — 11 letters with A:2, R:2, N:2, E:2, and G,M,T:1 each.
   2. Total distinct arrangements $=\dfrac{11!}{2!\,2!\,2!\,2!}=\dfrac{11!}{16}$.
   3. To count arrangements with the two E's adjacent, glue them into a single unit "EE" (they're identical, so no internal ordering needed). Now we arrange $10$ units: A,A,R,R,N,N,G,M,T,[EE] — repeats A:2,R:2,N:2 remain. Arrangements $=\dfrac{10!}{2!\,2!\,2!}=\dfrac{10!}{8}$. …

   🔒 Start your 14-day free trial to see the full solution →
9. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
   Q.If $2^n$ divides $16!$ and $2^{n+1}$ does not divide $16!$, then $n =$ (A) $14$ (B) $15$ (C) $16$ (D) $17$

   ›Reveal solutionSolution

   Legendre's formula sums $\lfloor 16/2^i\rfloor$ for $i=1,2,3,4$ to get the exact power of $2$ in $16!$, which is $15$.

   Concept and Intuition

   The exponent of a prime $p$ in $n!$ is given by Legendre's formula: $\sum_{i=1}^{\infty}\left\lfloor\dfrac{n}{p^i}\right\rfloor$ — this counts multiples of $p$, then multiples of $p^2$ (which contribute an extra factor), and so on, until $p^i>n$.

   Step-by-Step Solution

   1. Apply Legendre's formula with $n=16$, $p=2$: sum $\left\lfloor\frac{16}{2}\right\rfloor + \left\lfloor\frac{16}{4}\right\rfloor + \left\lfloor\frac{16}{8}\right\rfloor + \left\lfloor\frac{16}{16}\right\rfloor + \left\lfloor\frac{16}{32}\right\rfloor+\ldots$ …

   🔒 Start your 14-day free trial to see the full solution →
10. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
    Q.If $f(n) = n!(31-n)!$, where $n \in \{0,1,2,\ldots,31\}$, then the minimum value of $f(n)$ is (A) $(15!)(15!)$ (B) $(15!)(14!)$ (C) $(14!)(16!)$ (D) $(15!)(16!)$

    ›Reveal solutionSolution

    $f(n)=n!(31-n)!$ is inversely proportional to $\binom{31}{n}$, so it's smallest where the binomial coefficient is largest — at $n=15$ or $16$, giving minimum value $15!\cdot16!$.

    Concept and Intuition

    Binomial coefficients $\binom{N}{n}$ are largest at the middle value(s) of $n$ and taper off symmetrically toward the ends. Since $f(n)=n!(31-n)!=31!/\binom{31}{n}$, maximizing $\binom{31}{n}$ minimizes $f(n)$.

    Step-by-Step Solution

    1. Note $\binom{31}{n}=\dfrac{31!}{n!(31-n)!}=\dfrac{31!}{f(n)}$, so $f(n)=\dfrac{31!}{\binom{31}{n}}$.
    2. $f(n)$ is minimized exactly when $\binom{31}{n}$ is maximized.
    3. Since $31$ is odd, $\binom{31}{n}$ attains its maximum at the two central, equal values $n=15$ and $n=16$ (as $\binom{31}{15}=\binom{31}{16}$). …

    🔒 Start your 14-day free trial to see the full solution →
11. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
    Q.If $N(n)=n\displaystyle\prod_{r=1}^{2023}(n^2-r^2)\ (n>2023)$, then ${}^{N}C_{N-1}$ when $n=2024$ is (A) $(4047)!$ (B) $(4048)!$ (C) $(6023)!$ (D) $(6069)!$

    ›Reveal solutionSolution

    The product telescopes to $N(2024)=4047!$, and since $\binom{N}{N-1}=N$ for any $N$, the answer is simply $4047!$.

    Concept and Intuition

    Two ideas combine here: (1) the product $\prod_{r=1}^{k}(n-r)(n+r)$ telescopes into ratios of factorials, and (2) the combinatorial identity $\binom{N}{N-1}=N$ (choosing $N-1$ out of $N$ is the same as leaving one out, of which there are $N$ ways) means we never actually need to compute a huge binomial coefficient — just $N$ itself.

    Step-by-Step Solution

    1. $N(n) = n\displaystyle\prod_{r=1}^{2023}(n^2-r^2) = n\displaystyle\prod_{r=1}^{2023}(n-r)(n+r)$.
    2. At $n=2024$: as $r$ runs $1$ to $2023$, $(n-r)$ runs from $2023$ down to $1$, so $\prod(n-r)=2023!$.
    3. As $r$ runs $1$ to $2023$, $(n+r)$ runs from $2025$ to $4047$, so $\prod(n+r)=\dfrac{4047!}{2024!}$ (product of all integers up to 4047, divided by the product up to 2024). …

    🔒 Start your 14-day free trial to see the full solution →
12. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
    Q.A natural number $n$ such that $n!$ ends in exactly 1000 zeros is (A) $4010$ (B) $4000$ (C) $4009$ (D) $4004$

    ›Reveal solutionSolution

    Using Legendre's formula for the power of 5 dividing $n!$, only $n=4009$ (among the given choices) gives exactly 1000 trailing zeros; the count jumps to 1001 at $n=4010$ and sits at 999 for $n=4000$ or $4004$.

    Concept and Intuition

    The number of trailing zeros in $n!$ equals the exponent of $5$ in its prime factorisation (since factors of $2$ are always in surplus), computed by Legendre's formula: $Z(n)=\sum_{i=1}^{\infty}\left\lfloor \dfrac{n}{5^i}\right\rfloor$. This function only increases at multiples of $5$, and stays constant over each run of 5 consecutive integers between one multiple of 5 and the next.

    Step-by-Step Solution

    1. Compute $Z(4009) = \left\lfloor\frac{4009}{5}\right\rfloor+\left\lfloor\frac{4009}{25}\right\rfloor+\left\lfloor\frac{4009}{125}\right\rfloor+\left\lfloor\frac{4009}{625}\right\rfloor+\left\lfloor\frac{4009}{3125}\right\rfloor$ $=801+160+32+6+1 = 1000$.
    2. Compute $Z(4010) = 802+160+32+6+1 = 1001$ (jumps because $4010$ is again a multiple of 5, adding one more to the first term). …

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