how many sets of 3 in 4
nCr = n!/(r!(n-r)!)
4C3 = 4!/(3!(4-3)!)
= 24/(6*1)
= 4how many sets of 3 in 4
nCr = n!/(r!(n-r)!)
4C3 = 4!/(3!(4-3)!)
= 24/(6*1)
= 4Copyright © 2021 Codeinu
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