Combinations


Martin McBride, 2021-11-04
Tags combinations
Categories statistics probability

Combinations are similar to permutations. The difference is that, with combinations, the order of the items doesn't matter.

Example of combinations

One example of combinations is the National Lottery. The machine has 59 balls, numbered 1 to 59. In the game, 6 balls are chosen.

It doesn't matter which order the balls are drawn, if the 6 numbers match your ticket you will win. If the numbers are drawn in the order (10, 17 45, 23, 52, 36) the result is exactly the same as the numbers being drawn in the order (45, 52, 17, 36, 10, 23). The two situations are considered identical in terms of the final result.

In other words

  • (10, 17 45, 23, 52, 36) and (45, 52, 17, 36, 10, 23) are identical combinations, because each set contains the same numbers.
  • They are different permutations, because the numbers are in a different order.

Calculating the number of combinations

In the article on permutations, we used an example of 5 different coloured counters - red, yellow, green, blue, and black.

If we pick 3 of the counters at random, how many possible combinations are there?

We know that the number of permutations of 3 elements from 5 is:

$$ \frac {n!}{(n-r)!} = \frac {5!}{2!} = 60 $$

However, some of these different permutations are the same combination. For example the permutations RGB, GRB and BGR are all the same combination of a red, a green, and a blue counter.

In fact, for every combination of 3 counters, there are 3 factorial permutations. So to obtain the current number of combinations we must divide the number of permutations by 3 factorial (in more generally r factorial):

$$ \frac {n!}{r!(n-r)!} = \frac {5!}{3!.2!} = 10 $$

Symmetry

The number of combinations of r items from n is equal to the number of combinations of (n-r) items from n.

This can be seen intuitively, as follows. Using the previous example, whenever we take a combination of 3 counters, we leave 2 counters behind. For each unique combination of 3 counters, there is a corresponding unique combination of the remaining 2 counters. So the number of combinations of 3 counters from 5 is equal to the number of combinations of 2 counters from 5.

Alternatively we can show this from the equation:

$$ \frac {n!}{r!(n-r)!} $$

If we replace r with (n - r) we get:

$$ \frac {n!}{(n-r)!(n-(n-r))!} = \frac {n!}{(n-r)!r!} = \frac {n!}{r!(n-r)!} $$

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