Example: a12345bcd
Note: there are 26 letters and 10 digits available
1) If numbers and letters can be repeated
We have 4 positions for letters and 5 positions for numbers.
So, we can choose 1 out of 26 letters to first position, 1 out of 10 numbers for each of next 5 positions, and 1 out of 26 letters for each of last 3 positions.
In this way we have: "26\\cdot10\\cdot10\\cdot10\\cdot10\\cdot10\\cdot26\\cdot26\\cdot26=26^4\\cdot10^5 =45,697,600,000" variants.
2) If no repeating is allowed.
In this case amount of numbers and letters we can choose will decrease by 1 on each step. (If we already have 'a' in the plate, we can't put it again)
"26\\cdot10\\cdot9\\cdot8\\cdot7\\cdot6\\cdot25\\cdot24\\cdot23=10,850,112,000"
Other solution. We can say, that we choose 4 different letters and 5 different numbers. And than we arrange them on their places.
"\\begin{pmatrix}\n26\\\\4\n\\end{pmatrix}\\cdot4!\\cdot\n\\begin{pmatrix}\n10\\\\5\n\\end{pmatrix}\\cdot5!"
Actually, I would not recoment this solution. It looks like round trip. But may be more convenient in some other tasks.
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