Q99
ArithmeticPermutations, Combinations & Probability
Let ${}^nP_r = n(n-1)(n-2)\ldots(n-r+1)$, where $n, r$ are integers such that $1 \le r \le n$. If ${}^nP_4 = 5040$, find $n$.
- a.10
- b.12
- c.20
- d.16
Answer: (A) 10
${}^nP_4=n(n-1)(n-2)(n-3)=5040$, a product of four consecutive integers. Trying $n=10$: $10\times9\times8\times7=5040$, so $n=10$. Check against the other options: for $n=12$ the product is $12\times11\times10\times9=11880$, already too large, and $n=16$ or $20$ would be larger still. A handy fact is that $5040=7!$, and $7!=10\times9\times8\times7$ because $6\times5\times4\times3\times2=720=10\times9\times8$.