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