Permutations and Combinations

Permutations

• n! is equal to $n\times (n-1)\times (n-2)\times (n-3)...\times 2\times 1$
• When order is important
• Think of it as arranging items
• Notation is ${}^n{P}_r$
  • Where we are arranging r objects from n objects
• This equation can be written as:

$\frac{n!}{(n-r)!}$

${}^n{P}_n=\frac{n!}{(n-n)!}=\frac{n!}{0!}=n!$

• Therefore, 0! is equal to 1
• If we are replacing the items, then the equation is $n^r$
  • E.g., if we have 5 digits, and we make a code of 3 digits, using those 5 digits (with repeating allowed), the amount of codes we can make is $5^3$
• If we have 3 letters: A, B, C - We have 6 permutations of these letters
  • 3! = 6
  • If we want to write them out, we have ABC, ACB, BAC, BCA, CAB, CBA
  • Thus they have multiple permutations

If we are arranging n objects, and there are p repeating objects, q repeating objects and m repeating objects:

$Permutations=\frac{n!}{p!q!m!}$

Combinations

• Order is not important
• Here, we are selecting items
• Notation is ${}^n{C}_r$, or ${(}_r^n)$
  • We are selecting r objects from n objects
• This equation can be written as:

$\frac{n!}{r!(n-r)!}$

${}^n{C}_n=\frac{n!}{n!(n-n)!}=\frac{n!}{n!0!}=1$

• Again, this confirms that 0! is equal to 1
• If we have 3 letters: A, B, C - We have 1 combination of these letters
  • If we want to write them out, we have ABC, ACB, BAC, BCA, CAB, CBA
  • These all contain the same letters
  • Therefore, it only has 1 combination
• Can be used in probability
  • E.g., how many ways are there where out of 10 boys and 5 girls, 1 boy is selected?
  • $\frac{{}^{10}{C}_1}{{}^{15}{C}_1}=\frac{10}{15}=\frac{2}{3}$