Permutations & Combinations Probability

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

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

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.

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

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:

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! =

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

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

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

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?

Solution P(2 jacks) = 2 jacks from 4 jacks 2 cards from 52 cards = 4 C 2 = 6 = 1 52 C

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) = C 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