7.3 Permutation or Combination

4/12/2013

In the last lesson (combination) we learned about possible number of combinations where the order in which things are chosen does not matter. Today, we’re going to learn about possible number of combinations where the order in which things are selected DOES matter. It’s called PERMUTATION.Example: Suppose there are 7 swimmers entering a swim meet. Assuming that there are no ties, in how many ways could the gold, silver, and bronze medals be awarded?

You want to take Math, Science and Art during Periods 1, 2 and 3. How many possible arrangements can you have?

Period 1 Period 2 Period 3

1 Math Science Art 2 Math Art Science3 Science Math Art 4 Science Art Math 5 Art Math Science6 Art Science Math

Period 1 Period 2 Period 3

3 2 1● ● = 6

Permutationan ordered arrangement of a set of objects

nPk

nPk is the number of possible arrangements of n objects given k number of spots.

Make the number of spots, fill each spot with possible number of choices, then multiply together.

Definition:

How it is written:

Formula:

Another method:

You want to take Math, Science and Art during Periods 1, 2 and 3. How many possible arrangements can you have?

Using the formula:Since there are 3 objects and there are 3 spots: n = 3 and k = 33P3

There are 9 candidates in a city election for mayor. The winner will be the mayor and the runner-up will be the vice mayor. How many different cabinets of mayor and vice-mayor are possible?

Since order DOES matter, it’s a permutation problem.

9P2 =

Or

¿9 ∙8 ∙7… .3 ∙2 ∙17 ∙6…3 ∙2∙1= 72

Mayor Vice-Mayor

9 8● = 72

Suppose there are 7 swimmers entering a swim meet. Assuming that there are no ties, in how many ways could the gold, silver, and bronze medals be awarded?

Since order DOES matter, it’s a permutation problem.

7P3 =

Or

¿7 ∙6 ∙5 ∙ 4 ∙3 ∙2 ∙14 ∙3 ∙2∙1= 210 ways a medal could be awarded

Gold Silver

7 6● = 210Bronze

5●

In how many distinct ways can the word STUDY be written?

Since there are 5 letters and 5 spots: n = 5 and k = 5

5P5 =

Or

¿5 ∙4 ∙3 ∙2∙10 !=1= 120

1 2 3 4 5

5 = 120• 24 3 1• • •

There is a raffle to be held in a classroom of 24 students. If 5 prizes are to be raffled, how many sets of winning students could there be?

Since order DOES NOT matter, it’s a combination problem.

24C5 =

24C5 = ¿24 ∙23 ∙22∙21 ∙20 ∙19… .3 ∙2∙119∙18…3 ∙2 ∙1 ∙5 ∙4 ∙3 ∙2∙1

= 42504

Homework

WS 7.3Skip #2 & 4

I’m reading a book about anti-gravity. I can’t put it down!!!