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.

  • 1
    $\begingroup$ 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... $\endgroup$ – poncho Dec 6 '18 at 18:39

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