a 7 - digit number comprises of only 2 ' s and 3 ' s . how many of these are multiples of 12 ?
detailed solution number should be a multiple of 3 and 4 . so , the sum of the digits should be a multiple of 3 . we can either have all seven digits as 3 , or have three 2 ' s and four 3 ' s , or six 2 ' s and a 3 . ( the number of 2 ' s should be a multiple of 3 ) . for the number to be a multiple of 4 , the last 2 digits should be 32 . now , let us combine these two . all seven 3 ' s - no possibility . three 2 ' s and four 3 ' s - the first 5 digits should have two 2 ' s and three 3 ' s in some order . no of possibilities = 5 ! 3 ! 2 ! = 10 six 2 ' s and one 3 - the first 5 digits should all be 2 ' s . so , there is only one number 2222232 . so , there are a total of 10 + 1 = 11 solutions . correct answer : a .
add(divide(factorial(add(3, 2)), multiply(factorial(3), factorial(2))), const_1)