Question

Find the value of `""^80"C"_2`

Answer 1

`""^80"C"_2 = (80!)/(2!(80 - 2)!)` ...`(∵  ""^"n""C"_"r" = ("n"!)/("r"!("n" - "r"!)!))`

= `(80!)/(2!78!)`

= `(80 xx 79 xx 78!)/(2 xx 78!)`

= 40 × 79
= 3160