Question

Find the value of
1) ${}^5{p}_2{+}^4{p}_0{-}^3{p}_1$
2) ${}^{100}{C}_{100}{-}^{100}{C}_0$

Answer 1

1)${}^5{p}_2=\frac{5\times 4\times 3\times 2\times 1}{2\times 1}=60$
${}^4{p}_0=\frac{4\times 8\times 2\times 1}{0}=24$
${}^3{p}_1=\frac{3\times 2\times 1}{1}=06$
${}^5{p}_2{+}^4{p}_0{-}^3{p}_1$
= 60+6-24
= 66-24
= 36
2) ${}^{100}{C}_{100}{-}^{100}{C}_0$
${}^n{C}_r=\frac{n!}{(n-r)r!}$
$\therefore n=100,r=100$
$\therefore \frac{100}{(100-100)100}$
$=\frac{100}{0\left(100\right)}$
$=\frac{100}{0}=100$