Q.How many words with or without meaning can be made from the letters of the word MONDAY?

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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