What is the sum of <strong>ALL</strong> four digit numbers formed by using <strong>ALL</strong> digits 0, 1, 4, 5 without repetition of digits?

Explanation

Total permutations of 0,1,4,5 is 4! = 24. The sum of all these 24 numbers is 3! \times (0+1+4+5) \times 1111 = 6 \times 10 \times 1111 = 66660. We must subtract permutations starting with 0 (which are 3-digit numbers). Their sum is 2! \times (1+4+5) \times 111 = 2 \times 10 \times 111 = 2220. The valid sum is 66660 - 2220 = 64440.

Related questions on Algebra

• How many four-digit natural numbers are there such that <strong>ALL</strong> of the digits are odd?
• What is \sum_{r=0}^{n}2^{r}C(n,r) equal to ?
• If different permutations of the letters of the word 'MATHEMATICS' are listed as in a dictionary, how many words (with or without meaning) a...
• Consider the following statements : 1. If f is the subset of Z\times Z defined by f=\{(xy,x-y);x,y\in Z\}, then f is a function from...
• For how many quadratic equations, the sum of roots is equal to the product of roots?