Question:
Evaluate $\frac{n !}{(n-r) !^{2}}$, when
(i) n = 6, r = 2
(ii) n = 9, r = 5
Solution:
(i) When $n=6, r=2, \frac{n !}{(n-r) !}=\frac{6 !}{(6-2) !}=\frac{6 !}{4 !}=\frac{6 \times 5 \times 4 !}{4 !}=30$
(ii) When $n=9, r=5, \frac{n !}{(n-r) !}=\frac{9 !}{(9-5) !}=\frac{9 !}{4 !}=\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 !}{4 !}$
$=9 \times 8 \times 7 \times 6 \times 5=15120$