# How many permutations when each position takes from different length alphabets?

Let's say there's a password scheme as follows:

• the password is of length 4 (four)
• the 1st position is taken by one character from the set: a, b, c
• the 2nd position is taken by one character from the set: D, E, F, G
• the 3rd position is taken by one character from the set: 1, 2, 3, 4, 5
• the 4th position is taken by one character from the set: -, +

Example: b, F, 2, +

Is the number of permutations calculated as follows?

3C1 * 4C1 * 5C1 * 2C1 = 3 * 4 * 5 * 2 = 120

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Your answer is correct. –  Mhenni Benghorbal Dec 2 '12 at 7:03
If you have $n$ different "experiments" with $k_i$ outcomes each, $i = 1,2,...,n$ then the total number of outcomes is $$\prod_{i=1}^{n}k_i$$ To help you visualize this, think of just two experiments and list all of the outcomes from experiment 1. For each of these $k_1$ outcomes we can pair with it $k_2$ outcomes from the second experiment (giving $k_1k_2$ outcomes in total). The generalization comes by thinking of the first and second experiments as one experiment and inducting on the number of additional experiments. This property is of fundamental importance in combinatorics.