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Calculate combinations

Combinations count the distinct ways to select rr objects from nn distinct objects when order does not matter, represented by (nr)=n!r!(n−r)!\binom{n}{r}=\frac{n!}{r!(n-r)!} for 0≤r≤n0\le r\le n. The learner distinguishes combinations from permutations, understands why arrangements differing only in order are counted once, and applies symmetry (nr)=(nn−r)\binom{n}{r}=\binom{n}{n-r} in counting and probability contexts; selections with repetition and more abstract generalizations are outside this scope.

Detailed Explanation: Calculate combinations

Use a combination when you select objects and the order of selection does not matter. For example, choosing Alice, Ben, and Carla is the same group as choosing Carla, Alice, and Ben.

The number of ways to choose rr objects from nn distinct objects is

(nr)=n!r!(n−r)!.\binom{n}{r}=\frac{n!}{r!(n-r)!}.

The factors (r!)(r!) remove the repeated counting caused by arranging the same selected objects in different orders.

Example: A club has 88 members. How many different committees of 33 members can be chosen?

Since a committee is a group, not an ordered list, use a combination:

(83)=8!3!(8−3)!\binom{8}{3}=\frac{8!}{3!(8-3)!}

Simplify the factorials:

(83)=8!3!5!=8⋅7⋅6⋅5!(3⋅2⋅1)5!\binom{8}{3} =\frac{8!}{3!5!} =\frac{8\cdot7\cdot6\cdot5!}{(3\cdot2\cdot1)5!}

Cancel (5!)(5!):

(83)=8⋅7⋅63⋅2⋅1=3366=56\binom{8}{3} =\frac{8\cdot7\cdot6}{3\cdot2\cdot1} =\frac{336}{6} =56

Therefore, there are

56\boxed{56}

different committees.

Remember that combinations have the symmetry property

(nr)=(nn−r).\binom{n}{r}=\binom{n}{n-r}.

So, in this example, (83)=(85)\binom{8}{3}=\binom{8}{5}. Choosing 33 committee members is equivalent to choosing the 55 members who are not on the committee.

Learn by doing: Calculate combinations

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    Probability Permutation or Combination - Scenario to Value

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