# Combinations/Permutations
**URL:** https://boards.straightdope.com/t/combinations-permutations/481507
**Category:** Factual Questions
**Created:** [January 15, 2009, 10:55pm UTC](https://boards.straightdope.com/t/combinations-permutations/481507 "2009-01-15T22:55:30Z")
**Posts on this page:** 3
**Page:** 1
### Author:  [@Shai\_tan](https://boards.straightdope.com/u/Shai_tan)
#### Post date: [January 15, 2009, 10:55pm UTC](https://boards.straightdope.com/t/combinations-permutations/481507/1 "2009-01-15T22:55:30Z")
I want to calculate the number of combinations possible by pairing two items of different types.
For instance: letter/number pairs. If counting actual pairs without regard for order (A3 and 3A are the same pair and NOT counted twice) Is the formula:
# of letters multiplied by # of numbers = possible pairs?
Just double checking my assumption…
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### Author:  [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)
#### Post date: [January 15, 2009, 11:56pm UTC](https://boards.straightdope.com/t/combinations-permutations/481507/2 "2009-01-15T23:56:40Z")
Yes.
(Imagine putting the pairs into a rectangle, whose columns correspond to letters and whose rows correspond to numbers. You can count the number of cells by multiplying the number of columns by the number of rows)
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### Author:  [@Chessic\_Sense](https://boards.straightdope.com/u/Chessic_Sense)
#### Post date: [January 16, 2009, 12:07am UTC](https://boards.straightdope.com/t/combinations-permutations/481507/3 "2009-01-16T00:07:06Z")
The multiplication counting principle says that if there are A ways to make the first choice and B ways to make the second (and C ways…), then the combos are AxB(xC…).
You’ve got A letters and B numbers, so it’s AxB.