Question

The sum of all numbers greater than $1000$ formed by using the digits $0,1,2,3,$ no digit being repeated in any number is

• A. $38664$
• B. $48664$
• C. $58664$
• D. None of these

Answer 1

The correct option is A $38664$

All the numbers are of 4 digits and they do not have 0 in thousand's place.

The number of numbers having 0 at unit's place$=3!$ (as the other three places are to be filled by 1,2 or 3)

The number of numbers having 1 in unit's place$={{}^{2}P}_1\times {{}^{2}P}_2$ (as thousand's place can be filled by one of 2,3 and the remaining two places can be filled by the remaining two digits)

Similarly, the number of numbers having 2 or 3 in unit's place$={{}^{2}P}_1\times {{}^{2}P}_2$ in each case.

Thus the sum of digits in units place for all the numbers$=3!\times 0+{{}^{2}P}_1\times {{}^{2}P}_2\times 1+{{}^{2}P}_1\times {{}^{2}P}_2\times 2+{{}^{2}P}_1\times {{}^{2}P}_2\times 3=24$

Similarly the sum of digits in ten's and hundredth's place$=24$ each.

Now thousand's place can be filled by $1,2$ or $3$.

Number of numbers having 1 in thousand's place$=3!$ (as the other 3 places will be filled by 0,2,3)

Number of numbers having 2 or 3 in thousands place$=3!$ each

Sum of digits in thousand's place for all the numbers$=3!\times 1+3!\times 2+3!\times 3=36$

Required sum of all numbers$=36\times 1000+24\times 100+24\times 10+24\times 1=38664$