Question

The sum of all the $4$-digit distinct numbers that can be formed with the digits $1,2,2$and $3$is

• A. $26664$
• B. $122664$
• C. $122234$
• D. $22264$

Answer 1

The correct option is A

$26664$

Explanation for correct option:

The $4$-digit distinct numbers that can be formed with the digits $1,2,2$and $3$is

$1223,1232,1322,3122,3212,3221,2132,2312,2213,2231,2321,2123$

The sum of the $4$-digit distinct numbers is $=26664$

Alternatively,

Digits are $1,2,2$ and $3$

Total number of distinct numbers ${}^4\mathrm{P}_2=\frac{4!}{(4-2)!}=12$

Total number when

1. $1$ at unit place is $3$
2. $2$ at unit place is $6$
3. $3$ at unit place is $3$

So, for different number $1$ is at unit, tenth hundred and thousand place $3$ times each

Therefore, $3·1·\left(1+10+{10}^2+{10}^3\right)$

Similarly for different number $2$ is at unit, tenth hundred and thousand place $6$ times each

Therefore $6·2·\left(1+10+{10}^2+{10}^3\right)$

Similarly for different number $3$ is at unit, tenth hundred and thousand place $3$ times each

Therefore $3·3·\left(1+10+{10}^2+{10}^3\right)$

So the sum of the number is

$$\begin{gathered} =\left(1·3+2·6+3·3\right)\left({10}^3+{10}^2+10+1\right) \\ =\left(24\right)·\left(1111\right) \\ =26664 \end{gathered}$$

Hence, option (A) is correct.