Numbers formed by taking each of the digits 2,3,5,6 and 8 once?

[Greg Bernhardt]

What is the sum of all the numbers formed by taking each of the digits 2,3,5,6 and 8 once?

 

[BoulderHead]

haha, maybe later.

 

[jamesrc]

6399936

 

[Greg Bernhardt]

Originally posted by jamesrc
6399936

How'd you get that answer?

 

[redrogue]

Hmm...combinatorics.

2 3 5 6 8

There are: (5)*(5-1)*(5-2)*(5-3)*(5-4) or 5*4*3*2*1 = 120 permutations for the above set of numbers.

One set of numbers add up to: 2+3+5+6+8 = 24.

So, the sum of all numbers formed from 120 permutations is:

120*24 = 2880.

 

[jamesrc]

If I answered the right question, for all of the 5-digit numbers you can make, each digit appears 24 times in each column, so you can write:

24*(22222+33333+55555+66666+88888)=6399936

Edit: Oh, I thought it was the sum of the numbers made by the combinations, not the sum of their digits.

 

[Greg Bernhardt]

Originally posted by jamesrc
If I answered the right question, for all of the 5-digit numbers you can make, each digit appears 24 times in each column, so you can write:

24*(22222+33333+55555+66666+88888)=6399936

Edit: Oh, I thought it was the sum of the numbers made by the combinations, not the sum of their digits.

James gets the point!

Here is an alternative way:
4!*(2+3+5+6+8)*(11111) = 6,399,936