# Probability and Permutations

## Key Questions

• 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.

• 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.

• This key question hasn't been answered yet.
• 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.

• This key question hasn't been answered yet.

## Questions

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