College Algebra & Trigonometry, 2018a

316 CHAPTER 7. COMBINATORICS
7.3 Permutations and Combinations
We saw in the last section that, when working with permutations, the order is
always important. If we were choosing 3 people from a group of 7 to serve on a
committee with no assigned roles, the nature of the problem would change.
For example, if we were choosing 3 people from a group of 7 to serve on a committee
as president, vice-president and treasurer, the answer would be 7 P 3 = 210.
But - if we wanted to choose 3 people from a group of 7 with no assigned roles,
then some of the choices in the permutation would be the same.
In a permutation:
1st place: Alice 1st place: Bob
2nd place: Bob 2nd place: Charlie
3rd place: Charlie 3rd place: Alice
the two choices listed above would be considered as being different and would
be counted separately. In a “combination” in which the order of selection is not
important and there are no assigned roles, we must compensate for these extra
choices.
If we are choosing 3 people from a group of 7 to serve on a committee with no
assigned roles then we should consider that any selection from a permutation
that includes the same three people should only be counted once.
So, when we select the three people, we should consider how many different
ways there are to group them and then remove those extra choices. In this example,
we are choosing three people. Each group of three can be arranged in six
different ways 3! = 3 ∗ 2=6, so each distinct group of three is counted six times.
In order to find the actual number of choices we take the number of possible
permutations and divide by 6 to arrive at the actual answer:
7C 3 = 7 P 3
3!
= 7!
4! ∗ 3!