# Possible combinations of 8 digits, 6 upper case letters, 2 distinct numbers

Assume I want to find out how many combinations there are for a 8 digit word with 6 uppercase letters and 2 distinct numbers, e.g. A7BC9DEF, 1A0CRDEF, ... how many combinations are there?

My approach's solution is : $$26^6 \cdot 90 \cdot 8 \cdot 7$$

90 because there are 90 possible numbers with two distinct digits, i.e. $$01, 02, \ldots, 09, 10, 12, \ldots, 21, 23, \ldots, 98 \rightarrow 99 - 9$$ valid numbers)

$$8 \cdot 7$$ because the digits of the 90 possible numbers can be found at $$8 \cdot 7$$ different indices.

I had combinatorics a long time ago in school so I thought asking here for clarification might be a good idea. Thanks in advance!

It's $$26^6 \times 90 \times 8 \times 7 / 2$$; the $$/2$$ is there because having the digits appear in locations 1 and 3 is exactly the same as having the digits appear in locations 3 and 1, and hence there are only 45, not 90 distinct locations where the digits may appear.
This is correct. If you had an $$n$$ symbol pattern with $$t$$ decimal digits, no two digits equal, and uppercase letters from a $$K$$ letter alphabet,you would have
$$K^{n-t}\binom{10}{t}\binom{n}{t}$$ allowable patterns.
• The formula you gave is not correct (and doesn't agree with CombinatorsN00b's answer); you don't account for the fact that you can choose place the digits in $\binom{n}{t}$ different places within pattern... – poncho Dec 6 '18 at 18:39