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SETS AND COUNTING
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2. 6.4 Permutations and Combinations
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3. Combinations

4. Combinations
In many situations, one is interested in determining the number of ways of selecting r objects from a set of n objects without any regard to the order in which the objects are selected. Such a subset is called a combination.

5. Combinations
To derive a formula for determining the number of combinations of n objects taken r at a time, written
C(n, r) or
we observe that each of the C(n, r) combinations of r objects can be permuted in r! ways (Figure 19).
Figure 19

6. Combinations
Thus, by the multiplication principle, the product r ! C(n, r) gives the number of permutations of n objects taken r at a time; that is,
r ! C(n, r) = P(n, r)
from which we find
C(n, r ) =
or, using Equation (6),

7. Combinations

8. Example 8
Compute and interpret the results of (a) C(4, 4) and (b) C(4, 2).
Solution:
This gives 1 as the number of combinations of four distinct objects taken four at a time.
Recall that 0! = 1.

9. Example 8 – Solution
cont’d
This gives 6 as the number of combinations of four distinct objects taken two at a time.

10. Combinations
The next example shows that solving a counting problem often involves the repeated application of Equation (6) and/or (8), possibly in conjunction with the multiplication principle.

11. Applied Example 11 – Selecting Members of a Group
The members of a string quartet consisting of two violinists, a violist, and a cellist are to be selected from a group of six violinists, three violists, and two cellists.
a. In how many ways can the string quartet be formed?
b. In how many ways can the string quartet be formed if one of the violinists is to be designated as the first violinist and the other is to be designated as the second violinist?

12. Applied Example 11 – Solution
a. Since the order in which each musician is selected is not important, we use combinations.
The violinists can be selected in C(6, 2), or 15, ways; the violist can be selected in C(3, 1), or 3, ways; and the cellist can be selected in C(2, 1), or 2, ways.
By the multiplication principle, there are 15  3  2, or 90, ways of forming the string quartet.
b. The order in which the violinists are selected is important here. Consequently, the number of ways of selecting the violinists is given by P(6, 2), or 30, ways.

13. Applied Example 11 – Solution
cont’d
The number of ways of selecting the violist and the cellist remain, of course, 3 and 2, respectively.
Therefore, the number of ways in which the string quartet can be formed is given by 30  3  2, or 180, ways.

14. Combinations
Note:
The solution of Example 11 involves both a permutation and a combination. When we select two violinists from six violinists, order is not important, and we use a combination formula to solve the problem. However, when one of the violinists is designated as a first violinist, order is important, and we use a permutation formula to solve the problem.

15. Practice
p. 354 Self-Check Exercises #1 & 2