How do you find the binomial coefficient of #((100), (2))#?

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Answer 1 · 2017-11-28T13:47:19

The answer is #=4950#

Apply

#((n),(k))=(n!)/((n-k)!k!)#

where #n! =nxx(n-1)xx(n-2)xx.......3xx2xx1#

Therefore,

#((100),(2))=(100!)/((100-2)!(2!))#

#=(100!)/(98!xx2!)#

#=(100*99)/(1*2)#

#=50*99#

#=4950#