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Wize High School Grade 12 Pre-Calculus Textbook > Permutations and Combinations [BC/AB]

Permutations

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Factorial Symbol (!)

  • n!n! is pronounced "n factorial"
  • It means n×(n−1) ×(n−2)×...×3×2×1n\times\left(n-1\right)\ \times\left(n-2\right)\times...\times3\times2\times1

Examples to Memorize!

2!=2!=
2 x 1 = 2
3!=3!=
3 x 2 x 1 = 6
4!=4!=
4 x 3 x 2 x 1 = 24
5!=5!=
5 x 4 x 3 x 2 x 1 = 120

Factorial Properties

  • 0!=10!=1
  • 1!=11!=1
  • n!=n(n−1)!n!=n\left(n-1\right)!

Simplifying Factorials

12!9! 3!\frac{12!}{9!\ 3!}
=12×11×10×9!9! 3!=\frac{12\times11\times10\times\color{red}9!} {\color{red}9!\color{black}\ 3!}
=12×11×103!=\frac{12\times11\times10}{3!}
=12×11×103×2×1=\frac{12\times11\times10}{3\times2\times1}
=2×11×10=2\times11\times10
=220=220
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Permutations

Number of permutations of n different objects = Number of ways to arrange n different objects
1st spot choices×2nd spot choices×...×(n-1)th spot choices×nth spot choices\begin{array}{ccccccccc} &\text{1st spot choices}& \times& \text{2nd spot choices}& \times& ...&\times&\text{(n-1)th spot choices}&\times& \text{nth spot choices}\\ \end{array}




Permutation with no restrictions

The number of permutations of n different objects is n!\color{red}n\color{black}!

Permutations of parts of a group

The number of permutations of n different objects taken k at a time is
n!(n−k)!\frac{\color{red}n\color{black}!}{\left(\color{red}n\color{black}-\color{orange}k\color{black}\right)!}

Permutations with repetition

The number of ways to arrange n objects, where a of them are identical of type 1, b of them are identical of type 2, c of them are identical of type 3, ... is
n!a! b! c!\frac{\color{red}n\color{black}!}{\color{orange}a!\ b!\ c!}

Wize Tip
Use some variation of these permutation formulas if the question involves keywords like "arrange", "order", "line up".

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Example: Permutation

In how many ways can 5 women and 5 men stand in a straight line for a photo if

a) they can stand however they want?
There are 10 different people → The number of ways to arrange 10 different things in a line is 10!10!

b) the women must stand side-by-side, and the men must stand side-by-side?
Wize Tip
If the question involves blocks of objects → arrange within each block, then arrange the blocks themselves
There are 5 different women → The number of ways to arrange 5 different things in a line is 5!5!
There are 5 different men → The number of ways to arrange 5 different things in a line is 5!5!
Finally, we still need to arrange the 2 different groups → The number of ways to arrange 2 different things in a line is 2!2!

Therefore, there are 5! 5! 2!5!\ 5!\ 2!different such photo lineups.

c) the women must stand side-by-side?
There are 5 different women → The number of ways to arrange 5 different things in a line is 5!5!
Now we add in the men. There are 5 men + the group of women → The number of ways to arrange 6 different things in a line is 6!6!

Therefore, there are 5! 6!5!\ 6! different such photo lineups

d) the women and men must alternate?
There are 5 different women → The number of ways to arrange 5 different things in a line is 5!5!
There's 2 possibilities for this alternating arrangement: (the represents a spot for the men):
  • Case 1: W W W W W
  • Case 2: W W W W W
There are 5 different men → The number of ways to arrange 5 different things in a line is 5!5!
Once we have the men arranged, they will fill in these spots in order.
So, there are 5! 5!5!\ 5! different such photo lineups for each case 1.

Therefore, there are 5! 5!+5! 5!=2×5! 5!5!\ 5!+5!\ 5!=2\times5!\ 5! such photo lineups.
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Example: Permutations

10 patients are participating in a clinical trial. In how many different ways can you assign 4 different drugs to some of the 10 patients if

a) each patient can be assigned at most 1 drug?
Method 1
You can treat this like a permutation question where you want to arrange 4 of the 10 patients (once arranged, they can take the 1st, 2nd, 3rd, and 4th drug accordingly) → 10!(10−4)!=10!6!\frac{10!}{\left(10-4\right)!}=\frac{10!}{6!}.

Method 2
You can treat this like a fundamental counting principle question.
1st drug choices×2nd drug choices× 3rd drug choices×4th drug choices\text{1st drug choices}\times\text{2nd drug choices}\times\ \text{3rd drug choices}\times\text{4th drug choices}
=10×9×8×7=10\times9\times8\times7
We can simplify this:
=10×9×8×7×6!6!=\frac{10\times9\times8\times7\times\color{red}6!}{\color{red}6!}
=10!6!=\frac{10!}{6!}

b) there are no restrictions on the number of drugs each patient can take?
1st drug choices×2nd drug choices× 3rd drug choices×4th drug choices\text{1st drug choices}\times\text{2nd drug choices}\times\ \text{3rd drug choices}\times\text{4th drug choices}
=10×10×10×10=10\times10\times10\times10
=104=10^4
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Example: Permutation

In how many ways can 5 identical math textbooks, 4 distinct physics textbooks, and 3 identical psychology textbooks be arranged on a book shelf?

There are 5+4+3=12 objects in total.
5 of them are identical of type 1 (math), and 3 of them are identical of type 2 (psychology).
Using our formula, there are 12!5! 3!\frac{12!}{5!\ 3!} such arrangements.

Practice: Permutation

2 dogs, 3 cats, and 4 turtles are at a pet party. In how many ways can you arrange these pets in a line if the line must start and end with a cat?

Practice: Permutation w/ Repeated Objects

Find the number of different permutations of the letters in the word "MISSISSAUGA", if

a) there are no restrictions?
b) the A's must be side-by-side, and the I's must be side-by-side?
a) there are no restrictions?