You need to evaluate the binomial coefficient, using the next formula, such that:
`nCk = (n!)/(k!(n-k)!)`
Replacing 5 for n and 3 for k yields:
`5C3 = (5!)/(3!(5-3)!) => 5C3 = (5!)/(3!*2!)`
Notice that 5`! = 1*2*3*4*5 => 5! = 3!*4*5`
`5C3 = (3!*4*5)/(3!*2!) => 5C3 = (4*5)/(2!) => 5C3 = (4*5)/(1*2) = 10`
Hence, evaluating the binomial coefficient yields `5C3 = 10. `
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