Why is this a permutation and not a combination?
I have the solution for the following question. The problem I am experiencing is I do not understand how the answer is found.
The question as follows:
In how many different ways can $7$ identical objects be distributed between $3$ ordered boxes, box 1, box 2 and box 3? For how many of these distributions is there at least one object in each box?
The solution I have finds the following first: $$\binom92 = 36$$
However, when I put this into the calculator I get an answer of $18$
The final answer is: $$\binom62 = 15$$
When I also put the above into the calculator I get $12$...
The solutions are correct and the numbers you got are not that of the corresponding $^nP_r$ since generally $^nP_r\ge \ ^nC_r$. It seems like you got $$9\times2=18$$and$$6\times2=12$$ instead of the correct answers.
Note that $$\binom{n}{r}=\frac{n!}{r!(n-r)!}$$