Review of Counting Part 1
Notes
Options Multiply!
1) If you have to choose between m options and then choose between
n options, then altogether there are m · n total options.
If you have to choose between n1 options, then choose between n2
options, then choose between n3 options, and so on until last you
choose between nk options, then altogether there are n1 · n2 · n3 · · · nk
total options.
Example: To choose one order of ice cream with a topping, you have
to choose between 2 ice cream flavors and choose between 3
toppings. Althogether there are 2 · 3 = 6 total options.
Extra: Options multiply, but the word ”total” may remind us of
adding. That is because m · n = m + m + · · · + m (added n many
times).
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Review of Counting Part 1
Notes
Options Multiply!
2) There are n! many ways to order a set of n objects.
This number is also n · (n − 1) · (n − 2) · · · 2 · 1.
n!
many ways to choose and order k objects from a
(n − k)!
set of n objects.
This number is also n · (n − 1) · (n − 2) · · · (n − (k − 1)).
There are
Example: To choose and order 1st place, 2nd place, and 3rd place
prizes from a set of 28 contestants, there are
28!
= 28 · 27 · 26 = 19656 total options.
25!
Extra: Counting in this fashion is called a permutation.
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Choosing k objects from a set of n objects, without order
The number of ways to choose k objects from a set of n objects
(when
order does not matter) is called n choose k. It is written as
n
.
k
Example 1: You have 5 books. You can only fit 3 in your bag.
The
5
.
3
number of ways you can choose which 3 to put in your bag is
Example 2: You have 5 books. You can only fit 3 in your bag. You
have to leave 2 at home.
The number of ways you can choose which
5
2 to leave at home is
.
2
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Notes
More Examples - Counting Part 2
Notes
Example 3: There are four ways to choose 1 object from the set
{♣, ♦, ♥, ♠}. Since thereare
4 options for choosing 1 object from a
4
set of 4 objects, we have
= 4.
1
Example 4: Below is a list of all the possible ways that 2 objects are
choosen
from the set {♣, ♦, ♥, ♠}. There are six different ways. So
4
= 6.
2
♣, ♦
♦, ♥
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♣, ♥
♦, ♠
♣, ♠
♥, ♠
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Devoloping n choose k formula
Notes
There are two tasks when choosing and ordering k objects from a set
of n:
1. Choosing which k objects from a set of n.
2. Order the k objects that we chose.
Again,
n
.
k
2. Ordering k objects is written as k!.
1. n choose k is written as
From ”Counting Part 1,” this number is
n!
. Options multiply,
(n − k)!
so altogether
n
n!
· k! =
k
(n − k)!
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The general formula
Notes
Dividing both sides by k! gives the general formula for n choose k:
n
Formula for
k
n
n!
=
k
(n − k)! · k!
Revisiting previous examples:
5
5!
Example 1:
=
=
3
2! · 3!
5
5!
Example 2:
=
=
2
3! · 2!
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120
120
=
= 10
2·6
12
120
120
=
= 10
6·2
12
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