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

Solution:

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 $(6!/3!){\times }^4{\mathrm{C}}_1=480$.
(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 $\left[6!/(2!2!){\times }^4{\mathrm{C}}_2\right]$. Therefore, the required number of numbers is 480 + 1080$=$1560.