Question

The sum of all the numbers that can be formed by using all the digits 3, 2, 3, 4 is

• A. 39996
• B. 13332
• C. 19998
• D. None of these

Answer 1

Answer: (A)

39996

The general formula for $n$ digits is given by (where no digit is zero):
$(n-1)!\times$ $(Sum$ $of$ $all$ $digits)\times \mathrm{111....1}(n$ $times)$
But this is when all digits are distinct. If there is a repetition, we divide the above by the factorial of number of times a repetition has occurred:
2 can come in each of the 4 places in $\frac{3!}{2!}=3$ ways.
So, sum of all the $2^{\prime }s$ = $3\times (2000+200+20+2)=6666$
Similarly, sum of all $4^{\prime }s$ = $3\times (4000+400+40+4)=13332$
3 can come in each of the 4 places in $3!=6$ ways.
So, sum of all the $3^{\prime }s$ = $6\times (3000+300+30+3)=19998$
Therefore, total sum = $6666+13332+19998=39996$
Hence, (A) is correct.