Combinations part two

Say we have four differently coloured cubes and we want to find how many combinations involving two cubes there are.

There are 12 permutations. We can calculate this by having 4 ways of choosing the first cube and 3 ways of choosing the second and $4 \times 3 = 12$ . We can also calculate this by applying the formula for permutations $\frac{n!}{(n-r)!}$ We are choosing 2 cubes from 4 so n = 4 and r = 2 so $\frac{n!}{(n-r)!} = \frac{4!}{(4-2)!} = \frac{4!}{2!} = \frac{24}{2} = 12$ .

Here are the 12 permutations. Notice that each combination appears twice. For example blue and yellow.

To find the number of combinations from the number of permutations we simply divide the number of combinations by the number of ways of arranging two cubes. This is $2 \times 1 = 2$ . So we divide the number of permutations by 2 to get the number of combinations. $\frac{12}{2} = 6$ . There are 6 combinations.

The formula for combinations is the same as the formula for permutations divided by the number of ways to arrange the number of selections.

The formula for choosing r permutations from n things is $\frac{n!}{(n-r)!}$ . In order to calculate the number of combinations we divide by the number if ways to arrange r things in r places. This is r! So, the number of combinations is calculated by $\frac{\frac{n!}{(n-r)!}}{r!} = \frac{n!}{r!(n-r)!}$

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Combinations part two