Let S denote the set of 4-digit numbers a b c d such that abcd 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 \_\_\_\_

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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 \_\_\_\_

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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 .

Let S denote the set of 4-digit numbers a b c d such that a>b>c>d...

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