Question

The sum of all the numbers that can be formed by writing all the digits $3,2,3,4$ only once is

• A. $39996$
• B. $49996$
• C. $57776$
• D. None of these

Answer 1

The correct option is A $39996$

The number of numbers having 2 in unit's place $=\frac{3!}{2!}$ because the other 3 places are to be filled by 3,3,4 of which 2 are identical.

The number of numbers having 3 in unit's place $=3!$ because the other 3 places are to be filled by 2,3,4.

The number of numbers having 4 in unit's place $=\frac{3!}{2!}$ because the other 3 places are to be filled by 3,2,3 of which 2 are identical.

Thus the sum of digits in unit's place$=\frac{3!}{2!}\times 2+3!\times 3+\frac{3!}{2!}\times 4=36$

Similarly the sum of digits in other places is also 36 each. So the required sum is $=36\times 1000+36\times 100+36\times 10+36\times 1=39996$