I have three numbers $a,b$ and $c$

How many different additions can I have ?

$a + a + a = 3a$

$a + a + b = 2a + b$

However, $a + b + a =2a + b$ which is the same addition as above so I neglect it.

$b + b + b = 3b$

$a + b + c = a + b + c$

and so on , How many different sums can I have ? It's definitely not a permutation $3^3$ and not even a permutation with repetition and It is also definitely not $3!$

How many different additions and why ?


What you are looking for are combinations with repetition or combinations with replacement.

The formula for this is ${n+r-1\choose{r}}$, where $n$ is the number of objects you are choosing from, and $r$ is how many you are choosing.

In this case we have $n=3$ and $r=3$ so

${3+3-1\choose{3}} = {5\choose{3}} = 10$ combinations.


One combination is to use $1a, 1b,1c$. Thus, you have $1$ combination.

Another combination is to use $2a,\, 1b,\, 0c$ or $1a,\, 2b,\, 0c$ etc. (i.e. you use an element twice, another element once, and the other element not at all). So, you need all the arrangements, which are $3!=6$.

Following the same procedure the final way is to use $3a,\, 0b, \,0c$ or $0a,\, 3b,\, 0c$ etc. which gives just $3$ combinations.

So, all the possible combinations are $1+6+3=10.$


It could be determined by counting how many $a, b$ are selected from 0 to 3.

For examples, 2 'a's, 1 'b', 0 'c' 3 'a's, 0 'b', 0 'c' 1 'a's, 0 'b', 2 'c'

If a=0, we could choose b from 0 to 3. Thus, 4 cases are available. Because a and b are chosen, c is automatically determined.

If a=1, b could be from 0 to 2. ==> 3 cases.

If a=2, b could be either 0 or 1. ==> 2 cases.

If a=3, we don't have any choice. ==> 0 case.

Thus, 10 total cases are available.


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