12-1 and 12-2 Permutations and Combinations Get out that calculator!!

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12-1 and 12-2 Permutations and Combinations Get out that calculator!!

If we were to have a class trip to Great Adventure and there were only 3 rides (Ferris Wheel, Roller Coaster and Water Ride), in how many different orders could you ride them? 6 paths, right? You have 3 to choose from then 2, then one. (3 rides, picking one at a time)

Fundamental Principle of Counting If there are m 1 ways to do a first task, m 2 ways to do a second task, m 3 ways to do a third task…… m n ways to do the nth task, then the total possible “patterns” or ways you could do all the tasks is m 1 ·m 2 ·m 3...m n

Examples 1.How many 3 digit numbers can be formed if repetitions are allowed? 2.How many 3 digit numbers can be formed if repetitions are NOT allowed? 3.How many 4 letter “words” can be made if duplication of letters is allowed? 4.How many 4 letter “words” can be made if duplication of letters is not allowed?

Examples 5.How many 3 letter words can be formed in the first letter must be q? (no repetitions) 6.How many 3 letter words can be formed with middle letter m if repetitions is allowed? 7. In a game of poker, how many possible 5 card hands are there?

A new notation ! is called the factorial symbol n! = n(n-1)(n-2)(n-3)…..1 3! = 3(2)(1) = 6 5! = 5(4)(3)(2)(1)= 120

Permutations A permutation of “n” elements taken “r” at a time is an ordered arrangement (without repetition) of r of the n elements and it is called n P r. The thing to remember is that ORDER MATTERS!!

Examples 8. How many ways can 4 coins (dime, nickel, quarter and penny) be arranged in a row? 9. There is a club of 5 boys and 8 girls. The president will be a girl, the VP will be a boy, and the secretary and treasurer can each be either a boy or a girl. How many “councils” are possible??

What if order doesn’t matter? What if everyone in a class of 16 had to shake hands. Using the fundamental theorem of counting, don’t you get twice the number of handshakes? Because we would divide 16 P 2 by 2 since order doesn’t matter. When order doesn’t matter, it is called a Combination

Consider this question You and 3 of your friends go to Martinsville Pizza for an afternoon snack. You decide to get a large pie. Martinsville pizza offers 5 possible toppings (pepperoni, mushroom, sausage, green peppers and onion). How many possible pizza orders could be made? Keep in mind: You may have a pizza from 1 to 5 toppings.

Some shorthand P = pepperoni M = mushroom S = sausage G = green peppers O = onions

One topping P or M or S or G or O = 5 one topping pizzas No topping 1 no topping pizza

Two topping (Note: Sausage and Mushroom is the same as Mushroom and Sausage, right?) P with M, S, G or O (4) M with S, GP, or O (3) S with GP or O (2) GP with O (1) 10 two topping pizzas

Three topping P with M,S or M,G or M,O with S,G or S,O with G,O (6) M with S,G or S,O with G,O (3) S with G,O (1) 10 three topping pizzas

Four topping P with M,S,G or M,S,O or M,G,O or S,G,O (4) M with S,G,O (1) 5 four topping pizzas

Five topping There’s only one: P, M, S, G and O. 1 Five topping pizzas

So how many possible pizzas is this? 1+5+10+10+5+1 = 32 possible pizzas. Is there an easier way to determine these numbers without actually writing out the possibilities? This method seems very tedious – what if you are at an ice cream store and have 12 ice cream flavors and 9 toppings?

Lets just see if we can determine a formula Look at the 2 topping pizzas. At first, how many toppings can you choose from? How about the next number of choices? Look at this to see if it makes sense

P M S G O M S G O P S G O P M G O P M S O P M S G This looks like 20, which is 5 times 4. Why did we only have 10? Because we cancelled the overlap. There were twice as many pizzas because there were 2 sets of everything. Pepperoni/Onion is the same as Onion/Pepperoni.

Since there were two elements, for each combination, there was the matching one (in reverse order). So by dividing by the overlap, we got the true number of possibilities. What if there were 10 possible toppings and we were looking at two topping pizzas? 109 = 90 then divide by 2, so 45 possible pizzas.

What about other numbers of toppings? What if you did 3 toppings as if order mattered? What was the actual number of 3 topping pizzas? What if you did 4 toppings as if order mattered? What was the actual number of 4 topping pizzas? So what must we divide 5 P 3 by? What must we divide 5 P 4 by?

A Combination A Combination n elements, r at a time, is given by the symbol And can be expanded as

What does the formula do? determines the total number of possibilities (including overlap) and dividing by r! takes care of the overlaps. 2! = 2 (with 2 toppings) 3! = 6 (with 3 toppings)

Examples 10. I have a half dollar, penny, nickel, dime and quarter. How many 3 coin combinations are there? 11. I have a group of 10 boys, 15 girls. How many committees of 5 can I create of 2 boys and 3 girls? 12. How many teams of 6 hockey players be chosen from a group of 12 if position (i.e. order) doesn’t matter?

Homework p.745 11-25 odd q.752 11-17 odd

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