Question 891925
how many four digit numbers that are divisible by 3 can b formed using the digits 0,2,3,5,8, if no digit occurs more than once in each number?
<pre>
In order to be divisible by 3, the sum of the digits must be
divisible by 3, so we must selct 4 of those 5 digits that have
a sum which is divisible by 3. The sum of all 5 digits is 18,
so we can either leave out the 0 or the 3.  So,

we can select digits 0,2,5,and 8, since 0+2+5+8 = 15, which is divisible by 3.
Or we can select digits 2,3,5,and 8, since 2+3+5+8 = 18, which is divisible by
3.

If we select the digits 0,2,5,8

We can select the first digit any of 3 ways. (it can't be 0).
We can select the second digit any of 3 ways (it can be 0).
We can select the third digit either of 2 ways.
We can select the fourth digit 1 way.

That's 3*3*2*1 = 18 ways

If we select the digits 2,3,5,8

We can select the first digit any of 4 ways. 
We can select the second digit any of 3 ways.
We can select the third digit either of 2 ways.
We can select the fourth digit 1 way. 

That's 4*3*2*1 = 4! = 24

Total: 18+24 = 42 ways.

Edwin</pre>