1. Permutations & Combinations Probability

2. Warm-up How many distinguishable permutations are there for the letters in your last name?

3. Permutations An arrangement of a set of objects Example: Most bankcards can be used to access an account by entering a 4-digit PIN number. However knowing these 4 digits is not enough to access the account. The digits have to be in the correct order. Since the order is important, we must consider each arrangement as different

4. Combinations A selection from a group of objects without regard to order. If order were not important then any arrangement of the 4 numbers would access your bank account.

5. An expression for permutations The number of permutations of n objects taken r at a time reads “n permute r”

6. Permutations with repetitions The number of permutations of n objects, where a are the same of one kind, b are the same of another kind, and c are the same of yet another kind, can be represented by the expression:

7. Circular Permutations In general, the number of ways of arranging n objects around a circle is (n-1)! Example: At a graduation party, guests were seated in groups of 10 at circular tables. How many permutations are there for each table? (10-1)! = 9! = 362880

8. An expression for Combinations The number of combinations of n items taken r at a time reads: “n choose r”

9. Diagonals in a Polygon How many diagonals are there in a octagon? Choose 2 points out of 8 to be joined, then don’t count adjacent pairs. 20

10. Question A group of 4 journalists is to be chosen to cover a murder trial. There are 5 male and 7 female journalists available. How many possible groups can be formed: a)Consisting of 2 men and 2 women? b)Consisting of a least one woman? Solution a) 210b) 490

11. Application to Probability Remember that P(x) = # of favourable outcomes total number of outcomes Two cards are picked without replacement from a deck of 52 cards. What is the probability that both are jacks?

12. Solution P(2 jacks) = 2 jacks from 4 jacks 2 cards from 52 cards = 4 C 2 = 6 = 1 52 C 2 1362 221

13. The student council forms a sub-committee of 5 council members to look at how funds raised should be spent. If there are a total of 15 student council members, 6 males and 9 females, what is the probability that the sub-committee will consist of exactly 4 females? At least 4 females? P(4 females, 1male) = ( 9 C 4 )( 6 C 1 ) = (126)(6) = 36 15 C 5 3003 143 P(at least 4 females) = P(4 females) + P(5 females) = ( 9 C 4 )( 6 C 1 ) + ( 9 C 5 )( 6 C 0 ) = 3 15 C 5 15 C 5 11