Question

How do you differentiate between Permutations and Combinations in math? (Probability)

Answer 1

If you have
`n` objects and want to arrange them in a specific way that allows you distinguish one object from the other, then the arrangement is considered a permutation. The order in which the objects are arranged matters.

For example, here are all the permutations of a three digit number consisting of the digits 1, 2, 3, and 4:

123 213 312 412
124 214 314 413
132 231 321 421
134 234 324 423
142 241 341 431
143 243 342 432

We count this as


`{}_4P_3=(4!)/((4-3)!)=4!=24`

A combination *does not* take order into account. That's to say any two permutations that consist of the same objects will be considered the same arrangement.

In the above example, our combinations are

[123] 213 312 412
[124] 214 314 413
132 231 321 421
[134] [234] 324 423
142 241 341 431
143 243 342 432

In any number, we have 3 available digit positions. If we fixed the pool to, say, {1, 2, 3}, then we have
`3!` ways of arranging these numbers. But all of these arrangements are considered the same combination. So to count all the possible permutations of any 3 of the 4 digits, we could multiply the number of possible combinations of those same 3 digits by the number of ways we could arrange those 3 digits alone. In other words,
`3!{}_4C_3={}_4P_3`.

This immediately gives us a way to count the number of combinations.


`3!{}_4C_3={}_4P_3\implies{}_4C_3=\frac{{}_4P_3}{3!}=(4!)/(3!(4-3)!)=4`

More generally, given
`n` objects and taking
`k` at a time, the number of permutations is
`{}_nP_k=(n!)/(k!)` and the number of combinations is
`{}_nC_k=(n!)/(k!(n-k)!)`.

This comes up in probability when trying to determine the number of possible outcomes in the sample space for some given experiment. You have to be cautious about what the experiment is measuring and how it's being conducted.