How many combinations can you make with the numbers 1,2,3?

Mar 28, 2018

See a solution process below:

Explanation:

The first number in the combination can be any 1 of the 3 number.

The second number can be either of the 2 remaining numbers.

For the final number you would have only 1 choice.

Therefore, the number of combination is: $3 \times 2 \times 1 = 6$

1. $1 , 2 , 3$
2. $1 , 3 , 2$
3. $2 , 1 , 3$
4. $2 , 3 , 1$
5. $3 , 1 , 2$
6. $3 , 2 , 1$
Mar 28, 2018

Explanation:

We are not told the size of the combinations to be formed. I'll return to this point later.

Permutations and Combinations

In a permutation order matters, so the permutation (1 2 3) is not the same as (2 1 3).
In a combination order does not matter so the combination (2 3) is the same as (3 2).

Combinations with or without repetition

The question does not say whether we are allowed repetition or not.
So we do not know whether we are to count (1 1 3) as a combination of 3.

We are also not told the size of the combination. But if repetition is allowed then there are infinitely many combinations of all sizes that can be formed.
Therefore, I shall assume that repetition is not allowed.

Without repetition

Starting with 1 2 3 we can form combinations of size 1 2 or 3.
For $n$ things choosing $r$ combinations we can count using the formula

(n!)/(r!(n-r)!)

So we have:

$3$ choose $1$ in (3!)/(1!(3-1)!) = 3 ways
$3$ choose $2$ in (3!)/(2!(3-2)!) = 3 ways
$3$ choose $3$ in (3!)/(3!(3-3)!) = 1 way

That is a total of $7$ combinations. (If we wish to count choosing $0$ items -- the empty combination -- as a combination, then we must add $1$ way of doing that.)