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Permutations And Combinations

Question
CBSEENMA11014068

Evaluate:  fraction numerator straight n factorial over denominator straight r factorial space left parenthesis straight n minus straight r right parenthesis factorial end fraction comma when

(a) n = 6, r = 2,  (b) n = 6, r = 4, (c) n = 9, r = 4

Solution

(a) n = 6, r = 2
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(b)                n = 6, r = 4
rightwards double arrow           fraction numerator straight n factorial over denominator straight r factorial left parenthesis straight n minus straight r right parenthesis factorial end fraction equals fraction numerator 6 factorial over denominator 2 factorial space 4 factorial end fraction space equals space fraction numerator 6 cross times 5 cross times 4 factorial over denominator 4 factorial space 2 factorial end fraction space equals space fraction numerator 6 cross times 5 over denominator 2 cross times 1 end fraction equals 15
(c)               n = 9,  r = 4
rightwards double arrow          fraction numerator straight n factorial over denominator straight r factorial space left parenthesis straight n minus straight r right parenthesis factorial end fraction space equals space fraction numerator 9 factorial over denominator 4 factorial space 5 factorial end fraction equals fraction numerator 9 cross times 8 cross times 7 cross times 6 cross times 5 factorial over denominator 4 factorial space 5 factorial end fraction equals fraction numerator 9 cross times 8 cross times 7 cross times 6 over denominator 4 cross times 3 cross times 2 cross times 1 end fraction equals 126.      

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.