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Q.
The total number of natural numbers of six digits that can be made with digits $1,2,3,4$, if all digits are to appear in the same number at least once, is
Permutations and Combinations
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, that is, of the type 1,2 , $3,4,4,4$, etc. Number of ways of selection of digit which appears thrice $={ }^{4} C_{1}$
Then, number of numbers of this type
$=\frac{6 !}{3 !} \times{ }^{4} C_{1}=480$
(ii) Any two of the digits $1,2,3,4$ appears twice and the remaining two only once, that is, of the type $1,2,3$, $3,4,4$, etc.
Number of ways of selection of two digits which appears twice $={ }^{4} C_{2}$
Then, number of numbers of this type
$=\frac{6 !}{2 ! 2 !} \times{ }^{4} C_{2}=1080$
Therefore, the required number of numbers $=480+$ $1080=1560$