Q.If $\dfrac{1}{8!} + \dfrac{1}{9!} = \dfrac{x}{10!}$, find $x$.

We are given:

We need to find $n$ from the options.

Concept & Intuition

The ratio involves a combination (order doesn’t matter) and a permutation (order matters). Writing each in factorial form lets us cancel common factors and reduce to an algebraic equation. The key is to express both in terms of factorials, then simplify step by step, being careful with domain restrictions (e.g., $n-2 \ge 4$ so $n \ge 6$).

1. Write the combination and permutation in factorial form

2. Form the given ratio

Notice the $(n-6)!$ cancels neatly.

3. Set the ratio equal to $\frac{57}{16}$

4. Expand the factorial ratio

because $(n+2)! = (n+2)(n+1)n(n-1)(n-2)!$.

So the equation becomes:

5. Compute $8!$

Thus:

6. Cross-multiply

Compute $57 \times 40320$:

$57 \times 40000 = 2,280,000$

$57 \times 320 = 18,240$

Sum = $2,298,240$.

So:

7. Divide both sides by 16

8. Now we need a product of four consecutive integers equal to 143,640 …

Step 1 — What a bijection demands

$f:X\to Y$ is a bijection iff it is both

• injective (one‑one): distinct elements of $X$ have distinct images, and
• surjective (onto): every element of $Y$ is the image of some element of $X$.

Step 2 — The counting argument

If $f$ is injective, the image set $f(X)$ has exactly $|X|=7$ elements. If $f$ is also surjective, then $f(X)=Y$, so

But we are told $|Y| = 8$. Contradiction. Put plainly: 7 elements cannot be spread out to cover 8 targets — with only 7 arrows leaving $X$, at least one element of $Y$ is missed, so no function $X\to Y$ can be onto, let alone bijective.

The general rule this rests on:

$\text{Number of bijections } X\to Y=\begin{cases} n! & \text{if } |X|=|Y|=n\\[4pt] 0 & \text{if } |X|\neq |Y|\end{cases}$ …