Permutations And Combinations

Question
CBSEENMA11014099

Find the value of n such that:

space space fraction numerator straight P presuperscript straight n subscript 4 over denominator straight P presuperscript straight n minus 1 end presuperscript subscript 4 end fraction equals 5 over 3 comma space straight n greater than 4


Solution

space space space space straight P presuperscript straight n subscript 4 divided by straight P presuperscript straight n minus 1 end presuperscript subscript 4 space equals space 5 divided by 3
rightwards double arrow         bevelled fraction numerator space space space fraction numerator straight n factorial over denominator left parenthesis straight n minus 4 right parenthesis factorial end fraction over denominator begin display style fraction numerator left parenthesis straight n minus 1 right parenthesis factorial over denominator left parenthesis straight n minus 5 right parenthesis factorial end fraction equals space 5 divided by 3 end style end fraction


rightwards double arrow          fraction numerator straight n factorial over denominator left parenthesis straight n minus 4 right parenthesis factorial end fraction cross times fraction numerator left parenthesis straight n minus 5 right parenthesis factorial over denominator left parenthesis straight n minus 1 right parenthesis factorial end fraction equals 5 over 3
rightwards double arrow            space space space fraction numerator straight n left parenthesis straight n minus 1 right parenthesis factorial over denominator left parenthesis straight n minus 4 right parenthesis left parenthesis straight n minus 5 right parenthesis factorial end fraction cross times fraction numerator left parenthesis straight n minus 5 right parenthesis factorial over denominator left parenthesis straight n minus 1 right parenthesis factorial end fraction equals 5 over 3
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rightwards double arrow                             5n - 20 = 3n
rightwards double arrow                    2n = 20 or n = 10.

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Some More Questions From Permutations and Combinations Chapter

Determine K, so that K + 2, 4K – 6 and 3K – 2 are three consecutive terms of an A.P.