You need to evaluate the binomial coefficient, using the next formula, such that:
`((n),(k)) = (n!)/(k!(n-k)!)`
You should notice that `((n),(k)) = ((n),(n-k))`
Hence, `((100),(2)) = ((100),(100 - 2)) = ((100),(98))`
Evaluating `((100),(2))` yields:
`((100),(2)) = (100!)/(2!(100-2)!) => ((100),(2)) = (100!)/(2!(98)!)`
Notice that `100! = 98!*99*100`
`((100),(2)) = (98!*99*100)/(2!(98)!) => ((100),(2)) = (99*100)/(1*2) = 4950`
Hence, evaluating the binomial coefficient yields `((100),(98)) = 4950. `
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