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^nPr = ^nCr * ______________(a) r!(b) 1/r!(c) n!(d) 1/n!I had been asked this question by my school principal while I was bunking the class.My doubt stems from Combinations in portion Permutations and Combinations of Mathematics – Class 11

Answer»

The CORRECT choice is (a) r!

The best I can explain: We know, ^NPR = \(\FRAC{n!}{(n-r)!}\) and ^nCr = \(\frac{n!}{(n-r)! r!}\).

=> ^nPr / ^nCr = \(\frac{\frac{n!}{(n-r)!}}{\frac{n!}{(n-r)! r!}}\) = r!

=> ^nPr = ^nCr * r!



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