SOLUTION: "how many 7-digit numbers are there containing two 3s and three 8s if neither of the other two digits are 3 or 8( the first digit cannot be zero"

First we will allow the first digit to be any digit including 0 and 
then we'll subtract the number of ways the first digit is 0.

1.  Allowing the first digit to be any digit including 0.

We can choose the 2 positions to place the 2 3's in any of 7C2 or ways.
That leaves 5 positions for the 3 8's.
So we can choose the positions to place the 3 8's in any of 5C2 ways.
Each of the remaining 2 positions can be filled any of 8 ways 
{0,1,2,4,5,6,7,9}.

So that 8*8 or 64 ways to fill the remaining two positions. 

So that's (7C2)(5C2)(64)

2.  Now we must find the number of ways the 0 comes first to subtract
from that result.

Placing the 0 first leaves 6 positions to fill with digits.

We can choose the 2 positions to place the 2 3's in any of 6C2 or ways.
That leaves 4 positions to place the 3 8's.
So we can then choose the positions to place the 3 8's in any of 4C3 ways.
The 1 remaining position can be filled any of 8 ways {0,1,2,4,5,6,7,9},

So that's (6C2)(4C3)(8)

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So the final answer is (7C2)(5C2)-(6C2)(4C3)(8) = 13440-480 = 12960 ways.

Edwin