find the sum of all the distinct four digit numbers which have the property that their digits are among (1,2,3,4,5) and no digit appears twice in the same number
2006-09-07 13:09:05 · 2 answers · asked by vintagecrayon21 2 in Science & Mathematics ➔ Mathematics
It is an amusing question. Just a quick stab at it, but I think you'll find it is 24 * 16665 = 399960.
This is found, for me, by noting that the 5 occurs in the thousand's place 4P3 (or 24) times, as does the 4, the 3, the 2, etc. The same is true for the hundreds place, the tens place, and the ones place. So, you sum 5000, 4000, 3000,...500, 400, 300,...50, 40, 30,...2, 1. This yields 16665. This can then be multiplied by the permutation number to yield the end result, which agrees with the one stated above.
2006-09-07 14:43:13 · answer #1 · answered by galaxy625 2 · 0⤊ 0⤋
360*1111 or 399960
2006-09-07 21:33:13 · answer #2 · answered by Anonymous · 0⤊ 0⤋