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Permutations And Combinations

Question
CBSEENMA11014098
Wired Faculty App

Find the value of n such that:

straight P presuperscript straight n subscript 5 space equals space 42 straight P presuperscript straight n subscript 3 comma space straight n greater than 4

Solution
straight P presuperscript straight n subscript 5 space equals space 42 straight P presuperscript straight n subscript 3 comma space straight n greater than 4
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rightwards double arrow         fraction numerator 1 over denominator left parenthesis straight n minus 5 right parenthesis factorial end fraction space equals space 42. space fraction numerator 1 over denominator left parenthesis straight n minus 3 right parenthesis factorial end fraction

rightwards double arrow         fraction numerator 1 over denominator left parenthesis straight n minus 5 right parenthesis factorial end fraction space equals space 42. space fraction numerator 1 over denominator left parenthesis straight n minus 3 right parenthesis space left parenthesis straight n minus 4 right parenthesis space left parenthesis straight n minus 5 right parenthesis factorial end fraction
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rightwards double arrow           straight n squared minus 7 straight n plus 12 equals 42

rightwards double arrow            straight n squared minus 7 straight n minus 30 space equals 0
rightwards double arrow           (n - 10) (n + 3) = 0
rightwards double arrow                         n = 10                                 [∵ space space space straight n equals negative 3 not an element of straight N]
Hence,       n = 10

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

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