Q.$^{n}P_{r} =$

Showing the 12 most recent of 27 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $\dfrac{^nP_4}{^{n-1}P_4} = \dfrac{5}{3}$ then the value of $n$ is

   (a) 20

   (b) 2

   (c) 10

   (d) 1

   ›Reveal solutionSolution

   Write both permutation terms using factorials, cancel common factors, and solve the resulting linear equation.

   ${}^nP_4 = \dfrac{n!}{(n-4)!}, \qquad {}^{n-1}P_4 = \dfrac{(n-1)!}{(n-5)!}$ …

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2. CBSE 2026Set ANNUAL1 mark
   Q.${}^{n}P_r =$ ______, $(0 \le r \le n)$.

   ›Reveal solutionSolution

   $^{n}P_r = \dfrac{n!}{(n-r)!}$.

   The number of permutations of $n$ distinct objects taken $r$ at a time is given by choosing and arranging $r$ positions out of $n$ available objects without repetition: $n(n-1)(n-2)\cdots(n-r+1)$, which can be writt …

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3. CBSE 2025Set ANNUAL1 markMCQ
   Q.$^{11}P_0 =$

   (a) 0

   (b) 1

   (c) 11

   (d) 121

   ›Reveal solutionSolution

   $^{11}P_0 = 1$.

   The permutation formula is $^nP_r = \dfrac{n!}{(n-r)!}$.

   …

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4. CBSE 2025Set ANNUAL1 markMCQ
   Q.$^5P_0 - {}^4P_0 =$

   (a) 1

   (b) 0

   (c) -1

   (d) 9

   ›Reveal solutionSolution

   $^5P_0 - {}^4P_0 = 0$.

   Using $^nP_0 = 1$ for any $n$: $^5P_0 = 1$ and $^4P_0 = 1$.

   …

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5. CBSE 2025Set ANNUAL1 markMCQ
   Q.$^{12}P_2 =$

   (a) 121

   (b) 143

   (c) 132

   (d) 12

   ›Reveal solutionSolution

   $^{12}P_2 = 132$.

   $^nP_r = \dfrac{n!}{(n-r)!} = n(n-1)(n-2)\cdots(n-r+1)$, the product of $r$ consecutive integers starting from $n$.

   …

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6. CBSE 2025Set ANNUAL1 markMCQ
   Q.$^5P_3 =$

   (a) 60

   (b) 120

   (c) 180

   (d) 20

   ›Reveal solutionSolution

   $^5P_3 = 60$.

   $^nP_r = n(n-1)(n-2)\cdots(n-r+1)$.

   $^5P_3 = 5 \times 4 \times 3 = 60$. …

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7. CBSE 2025Set ANNUAL1 markMCQ
   Q.${}^nP_5 = 42 \times {}^nP_3$, $n > 4$ find $n$.

   (a) 12

   (b) 10

   (c) 14

   (d) 16

   ›Reveal solutionSolution

   Writing both permutation terms with factorials and simplifying the ratio reduces the equation to $(n-3)(n-4)=42$, whose solution (with $n>4$) is $n=10$.

   ${}^nP_5 = 42 \times {}^nP_3$

   $\frac{n!}{(n-5)!} = 42 \times \frac{n!}{(n-3)!}$

   Cancel $n!$ from both sides:

   $\frac{1}{(n-5)!} = \frac{42}{(n-3)!}$

   $\frac{(n-3)!}{(n-5)!} = 42$

   …

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8. CBSE 2025Set ANNUAL1 markMCQ
   Q.The number of 3 digit numbers formed by using digits 0, 1, 2, 3, 4, 5, when no digit is repeated, is:

   (a) 120

   (b) 100

   (c) 60

   (d) 720

   ›Reveal solutionSolution

   Total arrangements of 3 digits out of 6, minus the arrangements that start with 0 (not valid 3-digit numbers).

   Digits available: $0,1,2,3,4,5$ (6 digits), no repetition, forming 3-digit numbers.

   Total ways to arrange 3 digits chosen from 6 (ignoring the leading-digit restriction): $^6P_3 = 6\times5\times4 = 120$.

   …

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9. CBSE 2025Set sz1 markMCQ
   Q.The value of $\dfrac{n!}{(n-r)!}$, when $n = 5, r = 2$ is :

   (a) 20

   (b) 10

   (c) 30

   (d) 15

   ›Reveal solutionSolution

   $\dfrac{n!}{(n-r)!} = {}^{n}P_r$ counts ordered arrangements; for $n=5, r=2$ this is $5\times4=20$.

   By definition, ${}^nP_r = \dfrac{n!}{(n-r)!}$.

   Substitute $n=5$, $r=2$: …

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10. CBSE 2024Set ANNUAL1 markMCQ
    Q.Find the number of permutations of the letters of the word 'Independence'.

    (a) $\dfrac{12!}{3!\,2!\,4!}$

    (b) $\dfrac{9!}{3!\,4!\,2!}$

    (c) $\dfrac{12!}{1!\,3!\,5!}$

    (d) None of these

    ›Reveal solutionSolution

    Count the total letters and the repetition of each distinct letter, then apply the formula for permutations of a word with repeated letters: $\dfrac{n!}{p_1!\,p_2!\,\cdots}$.

    Write out INDEPENDENCE letter by letter: I, N, D, E, P, E, N, D, E, N, C, E — that's $12$ letters total.

    Count each distinct letter's frequency:

    • I: 1
    • N: 3 (positions 2, 7, 10)
    • D: 2 (positions 3, 8)
    • E: 4 (positions 4, 6, 9, 12)
    • P: 1
    • C: 1

    Check: $1+3+2+4+1+1 = 12$ ✓.

    The number of distinct permutations of a word with repeated letters is:

    $\frac{n!}{p_1!\,p_2!\,\cdots\,p_k!}$ …

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11. CBSE 2024Set ANNUAL1 markMCQ
    Q.If nP5 = 42 nP3, n > 4, the value of n is:

    (a) 10

    (b) 6

    (c) 0

    (d) 1

    ›Reveal solutionSolution

    $(n-3)(n-4)=42$ solves to $n=10$ (rejecting the invalid root $n=-3$).

    $^nP_5 = \dfrac{n!}{(n-5)!}$, $^nP_3 = \dfrac{n!}{(n-3)!}$.

    $\dfrac{^nP_5}{^nP_3} = \dfrac{(n-3)!}{(n-5)!} = (n-3)(n-4)$.

    Given $^nP_5 = 42\,^nP_3$:

    $(n-3)(n-4) = 42$

    $n^2 - 7n + 12 = 42$

    $n^2 - 7n - 30 = 0$ …

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12. CBSE 2024Set sz1 markMCQ
    Q.The value of $\dfrac{n!}{(n-r)!}$ when $n = 6, r = 2$ is equal to:

    (a) 6

    (b) 8

    (c) 15

    (d) 30

    ›Reveal solutionSolution

    $\dfrac{n!}{(n-r)!}$ with $n=6, r=2$ equals $30$.

    The expression $\dfrac{n!}{(n-r)!}$ is exactly the formula for $^nP_r$ (permutations of $n$ things taken $r$ at a time).

    Substitute $n=6, r=2$: …

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