Question

If \({}^nC_x=56\) and \({}^nP_x=336\), then find n and x.

• A. 7, 3
• B. 8, 4
• C. 8, 3
• D. 9, 6

Hint:

nPx / nCx = x!; use that to find x first, then solve n(n-1)(n-2) = 336 for n.

Answer 1

The Correct Option is C

Dividing the permutation value by the combination value directly isolates $x!$, which is usually the fastest route into these problems.

1. $\frac{{}^nP_x}{{}^nC_x} = x!$, so $x! = \frac{336}{56} = 6$, giving $x=3$ since $3!=6$.
2. With $x=3$, ${}^nC_3 = \frac{n(n-1)(n-2)}{6} = 56$, so $n(n-1)(n-2) = 336$.
3. Trying consecutive integers around a cube-root estimate of $336^{1/3} \approx 7$: testing $n=8$ gives $8 \times 7 \times 6 = 336$, a match.

So $n=8$ and $x=3$, confirming option C.