Fun with numbers! (Part-3)

sum all natural no 1-9 is 45. Since 7 digits selected sum up to a number divisible by three--->the other 2 no's are also divisible by 3.so we get 12 groups---(1,2)(1,5)(1,8)(2,7)(2,4)(3,6)(3,9)(4,5)(4,8)(5,7)(6,9)(7,8). the remaining 7 can be arranged in any manner so 12*(factorial{7}) is answer

We partition the digits 1,2,3,..,9 into congruence classes modulo 3, that is, we let A={3,6,9}, B={1,4,7} and C={2,5,8}. We can form the desired 7-digits by taking 1 element from A, 3 elements from B,3 elements from C, and 3 elements from A , 2 elements from B, 2 elements from C since the seven digit formed must be divisible by 3. Thus, there are $\dbinom{3}{1} x \dbinom{3}{3} x \dbinom{3}{3}+\dbinom{3}{3}x\dbinom{3}{2}x\dbinom{3}{2}=12$ 7-elements subset from {1,2,3,...,9} whose sum of the digits is divisible by 3. Hence the total number of 7-digits formed in these sets is $12 x 7!=60480$