If $\dfrac{1}{5!}+\dfrac{1}{6!}=\dfrac{x}{7!}$, then x equals to (a) 25 (b) 36 (c) 49 (d) 64

If (1)/(5!)+(1)/(6!)=(x)/(7!), then x equals to (a) 25 (b) 36 (c) 49 (d) 64

Question of 130

\dfrac{1}{5!}+\dfrac{1}{6!}=\dfrac{x}{7!}

(a) 25

(b) 36

(d) 64

Manipur CohsemCouncil of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2026MCQ· 1mImportance★★★★★

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Concept understanding — Factorial Arithmetic

Factorial Arithmetic — From Intuition to Precision

Imagine you have 3 different books you want to arrange on a shelf. How many different ways can you line them up? You could try listing them: Book A, B, C — or A, C, B — or B, A, C — and so on. If you actually count, you'll find 6 arrangements.

Where does that 6 come from? For the first position, you have 3 choices. Once you pick one, you have 2 choices left for the second position. Then only 1 choice remains for the last spot. So the total is $3 \times 2 \times 1 = 6$ .

That product — multiplying a whole number by every positive integer smaller than it, all the way down to 1 — is called a factorial. It's one of the most useful shortcuts in counting.

n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1

The symbol is an exclamation mark: $n!$ is read as "n factorial". It only makes sense for non-negative integers.

The first few values

$n$	$n!$	Why it matters
0	1	Special case (explained below)
1	1	Only one way to arrange one thing
2	2	Two ways: AB or BA
3	6	Three books, six arrangements
4	24	Four items, 24 arrangements
5	120	Grows fast — five books, 120 ways

Note

$0! = 1$ is not a guess — it's defined to make formulas work. There is exactly one way to arrange zero objects: do nothing. Also, many formulas like $n! = n \times (n-1)!$ would break at $n=1$ if $0!$ weren't 1.

The recursive nature

Factorials have a beautiful pattern: every factorial is the current number times the previous factorial.

5! = 5 \times 4!

4! = 4 \times 3!

3! = 3 \times 2!

2! = 2 \times 1!

1! = 1 \times 0! = 1 \times 1 = 1

This recursive definition is often how you'll compute factorials in problems: $n! = n \times (n-1)!$ , with the base case $0! = 1$ .

Why factorials explode so fast

Notice how quickly the numbers grow: $5! = 120$ , $6! = 720$ , $7! = 5040$ , $10! = 3,628,800$ . By $20!$ , you're already at 2.4 quintillion. This rapid growth is why factorials appear in probability (counting arrangements of decks of cards), combinatorics (choosing teams), and even in advanced mathematics like Taylor series.

Watch out

A common mistake: thinking $n!$ means $n$ multiplied by something else, like $n$ times some number. It's not — it's the product of all integers from $n$ down to 1. Also, factorials are not defined for negative numbers or fractions in basic arithmetic.

The core idea in one sentence …

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Solving 1/5! + 1/6! = x/7! gives x = 49.

Multiply every term by 7! = 7×6×5!:

$\dfrac{7!}{5!} + \dfrac{7!}{6!} = x$

…

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 19 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$\dfrac{1}{6!} + \dfrac{1}{7!} = \dfrac{x}{8!}$
(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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CBSE 2026Set ANNUAL1 markMCQ
$5!$
(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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CBSE 2025Set ANNUAL1 markMCQ
$2! =$
(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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CBSE 2025Set ANNUAL1 markMCQ
$3! + 1! =$
(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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CBSE 2025Set ANNUAL1 markMCQ
$4! - 3! =$
(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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CBSE 2025Set ANNUAL1 markMCQ
$\dfrac{1}{4!} + \dfrac{1}{5!} = \dfrac{x}{6!}$
(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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CBSE 2025Set ANNUAL1 mark
$5!$
›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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CBSE 2024Set ANNUAL1 markMCQ
$(n+1)! = 12\,(n-1)!$
(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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CBSE 2024Set ANNUAL1 markMCQ
$\dfrac{7!}{5!}$
(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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CBSE 2024Set ANNUAL1 markMCQ
$\dfrac{1}{9!} + \dfrac{1}{10!} = \dfrac{n}{11!}$
(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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CBSE 2023Set ANNUAL1 markMCQ
$\dfrac{1}{\lfloor 6} + \dfrac{1}{\lfloor 7} = \dfrac{x}{\lfloor 8}$
(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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CBSE 2023Set ANNUAL1 markMCQ
$\dfrac{8!}{6!\times 2!}$
(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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Q.If 15!+16!=x7!\dfrac{1}{5!}+\dfrac{1}{6!}=\dfrac{x}{7!}5!1​+6!1​=7!x​, then x equals to

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Q.If $\dfrac{1}{5!}+\dfrac{1}{6!}=\dfrac{x}{7!}$ , then x equals to

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