how to find all possible outcomes  

To find all possible outcomes of an event, we need to consider the concept of permutations and combinations, depending on the specific problem you're dealing with.
Permutations:
A permutation is an arrangement of objects in a specific order. The number of permutations can be found using the formula:
$P(n,r)=\frac{n!}{(n-r)!}$
where:
- $P(n,r)$ denotes the number of permutations of n objects taken r at a time.
- $n!$ represents the factorial of n, which is the product of all positive integers less than or equal to n.
- $(n-r)!$ represents the factorial of (n-r).
Combinations:
A combination is a selection of objects where the order doesn't matter. The number of combinations can be found using the formula:
$C(n,r)=\frac{n!}{r!(n-r)!}$
where:
- $C(n,r)$ denotes the number of combinations of n objects taken r at a time.
- $n!$ represents the factorial of n, as mentioned earlier.
- $r!$ represents the factorial of r.
- $(n-r)!$ represents the factorial of (n-r).
Now, let's go through an example to demonstrate how to find all possible outcomes using permutations and combinations.
Example:
Suppose we have a bag containing 4 colored balls: red, blue, green, and yellow. We want to find the number of ways we can arrange these balls if we pick 2 at a time.
To find the number of permutations, we use the formula $P(n,r)=\frac{n!}{(n-r)!}$. Plugging in the values, we get:
$P(4,2)=\frac{4!}{(4-2)!}=\frac{4!}{2!}=\frac{4·3·2·1}{2·1}=12$
Therefore, there are 12 different permutations of the 4 colored balls taken 2 at a time.
To find the number of combinations, we use the formula $C(n,r)=\frac{n!}{r!(n-r)!}$. Plugging in the values, we get:
$C(4,2)=\frac{4!}{2!(4-2)!}=\frac{4!}{2!·2!}=\frac{4·3·2·1}{2·1·2·1}=6$
Therefore, there are 6 different combinations of the 4 colored balls taken 2 at a time.