# Probability and Permutations

## Key Questions

• Permutations are arrangements of items, so the number of permutations is the number of arrangements of items.

Let $P \left(n , r\right)$ denote the number of permutations of $n$ items chosen $r$ items at a time. $P \left(n , r\right)$ can be found by

P(n,r)=n cdot (n-1) cdot (n-2)cdot cdots cdot(n-r+1)={n!}/{(n-r)!}.

Example

How many 3-digit codes are possible if each digit is chosen from 0 through 9, and no digits are repeated.

We can think of 3-digits codes as permutations of $10$ digits chosen $3$ digits at a time since no digits are repeated. So, we have

$P \left(10 , 3\right) = 10 \cdot 9 \cdot 8 = 720$.

Hence, there are 720 possible 3-digit codes.

I hope that this was helpful.

• For the first part of this answer, I will assume that the word has no duplicate letters.

To calculate the amount of permutations of a word, this is as simple as evaluating n!, where n is the amount of letters. A 6-letter word has 6! =6*5*4*3*2*1=720 different permutations.

To write out all the permutations is usually either very difficult, or a very long task. As you can tell, 720 different "words" will take a long time to write out. There are computer algorithms and programs to help you with this, and this is probably the best solution.

The second part of this answer deals with words that have repeated letters. One formula is
(n!)/(m_A!m_B!...m_Z!)
where $n$ is the amount of letters in the word, and ${m}_{A} , {m}_{B} , \ldots , {m}_{Z}$ are the occurrences of repeated letters in the word. Each $m$ equals the amount of times the letter appears in the word. For example, in the word "peace", ${m}_{A} = {m}_{C} = {m}_{P} = 1$ and ${m}_{E} = 2$. So the amount of permutations of the word "peace" is:
(5!)/(1!*1!*1!*2!) = (5*4*3*2*1)/(1*1*1*2*1) = 60

I will go through two more examples, but I will ignore every instance of 1! since 1! =1.

For the word "committee":
${m}_{C} = {m}_{O} = {m}_{I} = 1$
${m}_{M} = {m}_{T} = {m}_{E} = 2$
Permutations: (9!)/(2!2!2!) = (9*8*7*6*5*4*3*2*1)/((2*1)*(2*1)*(2*1)) = 45,360

For the word "cheese":
${m}_{C} = {m}_{H} = {m}_{S} = 1$
${m}_{E} = 3$
Permutations: (6!)/(3!) = (6*5*4*3*2*1)/(3*2*1) = 120

• It is important in permutations by the definition of the word "permutation". If we ignore order, then we are discussing / choosing "combinations".

Given set $\left\{a , b , c , d\right\}$, if we count $a b$ and $b a$ as the same, then we are counting combinations.
If they are counted as different, then we are counting permutation.

• Permutations of items are arrangements of items.

Examples

All six permutations of $\left\{a , b , c\right\}$ are:

$\left\{a b c , a c b , b a c , b c a , c a b , c b a\right\}$

Also, all 6 permutations of $\left\{a , b , c\right\}$ chosen at 2 items at a time are

$\left\{a b , b a , a c , c a , b c , c b\right\}$

I hope that this was helpful.