Question

The total number of $6$ digit numbers that can be made using digits $1,2,3,4$, if all the digits should appear in the number at least once is

• A. $1560$
• B. $840$
• C. $1080$
• D. $480$

Answer 1

The correct option is A $1560$

There can be two types of numbers.
$\left(i\right)$ 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 ${}^4{C}_1$.
Then the number of numbers of this type is $\left(\frac{6!}{3!}\right)\times {}^4{C}_1=480$

$\left(ii\right)$ 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 ${}^4{C}_2$.

Then the number of numbers of this type is $\frac{6!}{2!\cdot 2!}\times {}^4{C}_2=1080$

Therefore, the required number of numbers is $480+1080=1560$.