Three-digit numbers are formed from the digits 1,2 and 3 in such a way that the digits are not repeated. What is the sum of such three-digit numbers?

Step-by-step solution

Sum of the numbers = Sum of given numbers. ( n -1)! [10^0+10^1+10^2+ . . ] Here, sum of three digit numbers = Sum of the numbers (3-1)! [10^ +10^1+10^2 ] aligned & =(1+2+3)(3-1)! [10^0+10^1+10^2 ] & =6 2 1[1+10+100] & =12 111 & =1332 aligned