The total number of six-digit natural numbers that can be made with the digits 1 ,2, 3, 4, if all digits are to appear in the same number at least once is

There can be two types of numbers.(i) Any one of the digits 1, 2, 3, 4 appears thrice and the remaining digits only once, i.e., of the type 1, 2,3,4, 4, 4, etc. Number of ways of selection of digit which appears thrice is 4C1. Then the number of numbers of this type is (6!/3!)×4C1=480.(ii) Any two of the digits 1, 2, 3, 4 appears twice and the remaining two only once, i.e., of the type 1,2,3,3, 4, 4, etc. The number of ways of selection of two digits which appear twice is 4C2. Then the number of numbers of this type. is 6!/(2!2!)×4C2. Therefore, the required number of numbers is 480 + 1080=1560.