Download 7.2 Day 2 - Fremont Unified School District

Combinations, Permutations, and the
Fundamental Counting Principle
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If one event can happen m ways and another
event can happen n ways, the two events can
happen in sequence m x n ways.
You want to know how many different ice
cream sundaes you can make with the
following criteria:
◦ 1 flavor of ice cream (selected from 6 flavors)
◦ 1 flavor of syrup (selected from 3 flavors)
◦ 1 type of topping (selected from 8 choices)
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6*3*8=
144
There are 144 different ice cream sundaes
possible, choosing one item from each
category.
If you wanted to choose more than one off a
list, you would need to use “combinations”.
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Factorial is denoted by !
This represents the number of different
ordered arrangements of n distinct objects
(n!)
You multiply the numbers in descending
order until you reach 1:
◦ Example: 5! = 5 * 4 * 3 * 2 * 1 = 120
◦ Example: 3! = 3 * 2 * 1 = 6
◦ Rule for factorials: 0! = 1 (Special Case)
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How many ways items can be selected from a
larger group
Order does not matter
nCr:
n = total number in data set
r = number of items taken at a time
r≤n
r, n ≥ 0, must be integers
n
Cr = n! / (n-r)!r!
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Example: Use combinations for items on a pizza
(order of pepperoni, sausage, etc. doesn’t matter).
How many different 3 topping pizzas can be made if
there are 12 different toppings to choose from?
C3 = 12! / (12-3)! 3! =
12 * 11 * 10 * 9! / 9! * 3 * 2 * 1 =
12 * 11 * 10 / 3 * 2 * 1 = 220
12
There are 220 different 3 topping pizzas possible.
**Note about 9!
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6 types of ice cream (I want 2 different ones)
3 types of syrup (I still only want 1 kind)
8 types of toppings (I want 3 different ones)
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6
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C2 * 3C1 * 8C3 =
6C2 = 6! / (6-2)! 2! = 6*5*4! / 4!*2*1 = 15
3C1 = 3! / (3-1)! 1! = 3*2! / 2!*1 = 3
8C3 = 8! / (8-3)! 3! = 8*7*6*5! / 5!*3*2*1 = 56
So 15 * 3 * 56 = 2520
There are 2,520 different sundaes possible with the
above criteria.
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How many ways items can be chosen from a
larger group
Order does matter
nPr = n! / (n-r)!
There will be a larger number of
permutations than combinations.
Combination locks are actually permutation
locks.
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How many ways can 10 people running a race
finish 1st, 2nd, and 3rd?
10P3 = 10! / (10-3)! = 10 * 9 * 8 * 7! / 7!
= 10 * 9 * 8 = 720
There are 720 ways 10 people running a race
can finish 1st, 2nd, and 3rd.
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If the same number or letter is in a sequence,
the sequence may look the same even if you
have rearranged the digits/letters. To
determine how many distinguishable
permutations there are:
n! / n1!*n2!*n3!... where the denominator
terms are the repeats, such as this…
MISSISSIPPI: 11! / 4! * 4! * 2! =
◦ 4! = 4 I’s
◦ 4! = 4 S’s
◦ 2! = 2 P’s
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11*10*9*8*7*6*5*4! / 4!*4*3*2*1*2*1 =
11 * 10 * 3 * 7 * 3 * 5 =
110 * 21 * 15 =
34,650 ways
If we didn’t do “distinguishable” ways, there
are 39,916,800 permutations for arranging
the letters of the name MISSISSIPPI.
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See Schoolwires for a homework page with
problems related to these topics.
See me for additional copies.
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White Book: Page 157-158: #12-28 even
(There are fundamental counting principle,
permutation, and combination problems
here. If you just divide them up, you
probably won’t get practice with all the
concepts.)
We still have part 2 of the lesson to do…