Evaluate \(\displaystyle\frac{n!}{(n-r)!}\) when \(n=9\) and \(r=5\).
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 15120
\(\frac{n!}{(n-r)!}\) is \({}^{n}P_{r}\), the number of arrangements of \(r\) things out of \(n\). The factors of \((n-r)!\) cancel, leaving the product of the top \(r\) numbers.
With \(n=9\) and \(r=5\), the factors from 4 down to 1 cancel: \(\displaystyle\frac{9!}{4!}=9\times8\times7\times6\times5\)
In two groups: \(9\times8\times7=504\) and \(6\times5=30\)
Product of the two groups: \(504\times30=15120\)