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: Name the expressions.
$\frac{n!}{2!(n-2)!} = \binom{n}{2}$ and $\frac{n!}{4!(n-4)!} = \binom{n}{4}$. Their ratio is $2:1$.

Step 2: Write the ratio as a fraction.
\[ \frac{\binom{n}{2}}{\binom{n}{4}} = 2 \]
Step 3: Expand both combinations.
\[ \frac{\frac{n(n-1)}{2}}{\frac{n(n-1)(n-2)(n-3)}{24}} = 2 \]
Step 4: Cancel the common part.
Cancel $n(n-1)$ from top and bottom.
\[ \frac{12}{(n-2)(n-3)} = 2 \quad\Rightarrow\quad (n-2)(n-3) = 6 \]
Step 5: Solve the equation.
\[ n^2 - 5n + 6 = 6 \quad\Rightarrow\quad n^2 - 5n = 0 \quad\Rightarrow\quad n(n-5) = 0 \] So $n = 0$ or $n = 5$.

Step 6: Pick the valid value.
We need $n \ge 4$ for $\binom{n}{4}$ to make sense, so $n = 5$. \[ \boxed{n = 5 \text{ (Option 3)}} \]