Question

If $\frac{n!}{2!(n-2)!}$ and $\frac{n!}{4!(n-4)!}$ are in the ratio $2:1$, then $n =$

• A. $6$
• B. $4$
• C. $5$
• D. $3$

Hint:

For multiple-choice questions involving factorials or combinations, simply plugging the options into the original equation (e.g., trying $n=5$ first) is often much faster than solving the full quadratic equation algebraically.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a ratio involving two combination formulas (binomial coefficients) and asked to solve for the integer value $n$.

Step 2: Detailed Explanation:
The given expressions are the standard formulas for combinations:
$\frac{n!}{2!(n-2)!} = \binom{n}{2}$
$\frac{n!}{4!(n-4)!} = \binom{n}{4}$
The problem states their ratio is $2:1$, which can be written as a fraction:
$$\frac{\binom{n}{2}}{\binom{n}{4}} = \frac{2}{1}$$ Expand the combination formulas:
$$\frac{\frac{n(n-1)}{2}}{\frac{n(n-1)(n-2)(n-3)}{24}} = 2$$ Cancel the common terms $n(n-1)$ from the numerator and denominator:
$$\frac{\frac{1}{2}}{\frac{(n-2)(n-3)}{24}} = 2$$ Multiply by the reciprocal:
$$\frac{1}{2} \times \frac{24}{(n-2)(n-3)} = 2$$ $$\frac{12}{(n-2)(n-3)} = 2$$ Cross-multiply and isolate the polynomial:
$$12 = 2(n-2)(n-3)$$ $$6 = (n-2)(n-3)$$ Expand the right side:
$$6 = n^2 - 3n - 2n + 6$$ $$6 = n^2 - 5n + 6$$ Subtract 6 from both sides:
$$n^2 - 5n = 0$$ Factor out $n$:
$$n(n - 5) = 0$$ This gives possible solutions $n = 0$ or $n = 5$. Since $\binom{n}{4}$ requires $n \ge 4$ for the factorial to be defined, $n$ must be 5.

Step 3: Final Answer:
The value of $n$ is 5, matching option (C).