Q.If $\dfrac{1}{\lfloor 6} + \dfrac{1}{\lfloor 7} = \dfrac{x}{\lfloor 8}$ (factorial notation, $\lfloor n = n!$), then find $x$.

Showing the 12 most recent of 19 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $\dfrac{1}{6!} + \dfrac{1}{7!} = \dfrac{x}{8!}$ then value of $x$ is

   (a) 56

   (b) 64

   (c) 48

   (d) None of these

   ›Reveal solutionSolution

   Multiply both sides by $8!$ and simplify each factorial ratio using $n! = n\times(n-1)!$.

   $\dfrac{1}{6!} + \dfrac{1}{7!} = \dfrac{x}{8!}$

   Multiply both sides by $8!$:

   $\dfrac{8!}{6!} + \dfrac{8!}{7!} = x$ …

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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.The value of $5!$ is —

   (a) $5040$

   (b) $720$

   (c) $120$

   (d) $24$

   ›Reveal solutionSolution

   $5! = 120$, option (c).

   By definition, $n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1$.

   …

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3. CBSE 2025Set ANNUAL1 markMCQ
   Q.$2! =$

   (a) 1

   (b) 2

   (c) 4

   (d) 8

   ›Reveal solutionSolution

   $2! = 2$.

   By definition, $n! = n \times (n-1) \times \cdots \times 2 \times 1$.

   …

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4. CBSE 2025Set ANNUAL1 markMCQ
   Q.$3! + 1! =$

   (a) 6

   (b) $4!$

   (c) 7

   (d) $8!$

   ›Reveal solutionSolution

   $3! + 1! = 7$.

   $3! = 3 \times 2 \times 1 = 6$. $1! = 1$.

   $3! + 1! = 6 + 1 = 7$. …

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5. CBSE 2025Set ANNUAL1 markMCQ
   Q.$4! - 3! =$

   (a) $1!$

   (b) $3 \times 3!$

   (c) 16

   (d) $2 \times 3!$

   ›Reveal solutionSolution

   $4! - 3! = 18 = 3 \times 3!$.

   $4! = 24$, $3! = 6$. So $4! - 3! = 24 - 6 = 18$.

   Checking the options against 18: $1! = 1$ (no); $3 \times 3! = 3 \times 6 = 18$ (yes); $2 \times 3! = 2 \times 6 = 12$ ( …

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6. CBSE 2025Set ANNUAL1 markMCQ
   Q.If $\dfrac{1}{4!} + \dfrac{1}{5!} = \dfrac{x}{6!}$ then value of $x$ is

   (a) 40

   (b) 36

   (c) 38

   (d) 32

   ›Reveal solutionSolution

   Rewrite $\dfrac{1}{4!}$ and $\dfrac{1}{5!}$ with common denominator $6!=720$ and add; the numerator sum gives $x=36$.

   $\frac{1}{4!} + \frac{1}{5!} = \frac{x}{6!}$

   $4! = 24$, $5! = 120$, $6! = 720$.

   $\frac{1}{24} + \frac{1}{120} = \frac{30}{720} + \frac{6}{720} = \frac{36}{720}$

   So $\dfrac{x}{720} = \dfrac{36}{720} \implies x = 36$.

   …

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7. CBSE 2025Set ANNUAL1 mark
   Q.Fill in the blank: The value of $5!$ is ____.

   ›Reveal solutionSolution

   $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.

   By definition, $n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1$.

   …

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8. CBSE 2024Set ANNUAL1 markMCQ
   Q.If $(n+1)! = 12\,(n-1)!$, the value of $n$ will be

   (a) $n=3$

   (b) $n=4$

   (c) $n=50$

   (d) None of these

   ›Reveal solutionSolution

   Expand $(n+1)!$ in terms of $(n-1)!$ so the factorial cancels, leaving a simple quadratic in $n$.

   We are given $(n+1)! = 12(n-1)!$.

   Write $(n+1)! = (n+1)\cdot n \cdot (n-1)!$. So:

   $(n+1)\cdot n \cdot (n-1)! = 12(n-1)!$

   Since $(n-1)!\ne 0$, divide both sides by $(n-1)!$:

   $n(n+1) = 12$

   $n^2 + n - 12 = 0$

   $(n+4)(n-3) = 0$

   …

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9. CBSE 2024Set ANNUAL1 markMCQ
   Q.The value of $\dfrac{7!}{5!}$ will be —

   (a) 36

   (b) 42

   (c) 40

   (d) 24

   ›Reveal solutionSolution

   $\dfrac{7!}{5!}=42$, since the $5!$ in the denominator cancels with part of $7!$.

   …

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10. CBSE 2024Set ANNUAL1 markMCQ
    Q.If $\dfrac{1}{9!} + \dfrac{1}{10!} = \dfrac{n}{11!}$, then the value of $n$ is

    (a) 120

    (b) 125

    (c) 121

    (d) 130

    ›Reveal solutionSolution

    Express $\dfrac{1}{9!}$ and $\dfrac{1}{10!}$ with denominator $11!$ and add.

    Since $10! = 10\times 9!$ and $11! = 11\times 10\times 9! = 11\times 10!$:

    $\dfrac{1}{9!} = \dfrac{10\times 11}{11!} = \dfrac{110}{11!}$

    $\dfrac{1}{10!} = \dfrac{11}{11!}$

    …

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11. CBSE 2023Set ANNUAL1 markMCQ
    Q.If $\dfrac{1}{\lfloor 6} + \dfrac{1}{\lfloor 7} = \dfrac{x}{\lfloor 8}$ (factorial notation, $\lfloor n = n!$), then find $x$.

    (a) 54

    (b) 44

    (c) 74

    (d) 64

    ›Reveal solutionSolution

    Multiplying through by $8!$ converts the factorial fractions into whole numbers, giving $x=64$.

    We're given (using $\lfloor n$ to mean $n!$):

    $\frac{1}{6!}+\frac{1}{7!} = \frac{x}{8!}$

    Multiply every term by $8!$:

    $\frac{8!}{6!} + \frac{8!}{7!} = x$

    …

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12. CBSE 2023Set ANNUAL1 markMCQ
    Q.The value of $\dfrac{8!}{6!\times 2!}$ will be:

    (a) $28$

    (b) $64$

    (c) $56$

    (d) $18$

    ›Reveal solutionSolution

    $\dfrac{8!}{6!\times2!} = 28$.

    $\frac{8!}{6!\times2!} = \frac{8\times7\times6!}{6!\times2!} = \frac{8\times7}{2!} = \frac{56}{2} = 28$

    …

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