Let S denote the set of 4-digit numbers a b c d such that a>b>c>d and P denote the set of 5 -digit numbers having product of its digits equal to 20. Then n( ~S )+n( P ) is equal to _ _ _ _

Step-by-step solution

n(S) : Choose 4 distinct digits from 0,1, ,9 and arrange in decreasing order. n(S) = 10 4 = 210 . n(P) : 5-digit numbers with digit product = 20 = 2^2 5 . All digits 1 (else product = 0 ). Must include exactly one 5 , remaining 4 digits multiply to 4 . Case 1: 1,1,1,4,5 : 5! 3! = 20 arrangements. Case 2: 1,1,2,2,5 : 5! 2! 2! = 30 arrangements. n(P) = 50 . Hence n(S) + n(P) = 210 + 50 = 260 .