There are $n$ dogs and $k$ cats. In how many ways can we arrange them in a row so there are no $2$ cats are adjacent?
I thought about trying to calculate the number of possibilities without
any constraints and subtract the number of ways we can arrange them so there
is at least $1$ pair of adjacent cats.
Would like to understand how to approach this problem correctly (and if there are multiple ways I would love to hear them).
Edit: Sorry, I didn't make myself clear.
What I meant was that the cats and dogs are different(each dog is different
than the other and so are the cats).
And the row doesn't have to start with dogs – the only requirement is
that we can't have $2$ or more cats in a row.
But the other question (if the cats and dogs are the same) is good too.
If you have more ways to solve in case they are indistinguishable I would love
to read and understand it too.
So, basically this 1 question turned to two:
Case 1: the cats are indistinguishable and the dogs too.
Case 2: they are distinguishable.
And $n$ is large enough for it to be possible.
What is the minimal $n$? A bit confused.
If we start with a cat, then $n=k-1$ seems enough (cat first, cat last).
If we start with a dog, $n=k$. So in general we need $n$ at least $k$ to arrange them properly? Am I right?
Again sorry for the confusion!
Edit 2: My 2 questions were answered – thank you!
If you have any other ways to approach the problem or wish to add anything –
please do, I would gladly read it!
In particular, if you know how to place the cats first and then the dogs,
I would like to hear it.
Best Answer
Okay now, to solve this, we need a couple of constraints, the minimum value of $n$ has to be $n=k-1$ if it is more then also it doesn't make a difference. Now, let us say we keep the $n$ dogs in a row with space between each one. So clearly there will be $n+1$ spaces, now these $n+1$ spaces have to be occupied by $k$ cats. Thus we get $$\binom {n+1}{k}$$ This is assuming that the cats are all identical as are the dogs.
If not, then we get, $$\binom {n+1}{k}\cdot n! \cdot k!$$ The reason this works is to let us say, take an example where $k=5$ and $n=5$. Now we put the $5$ dogs in a line with space between them. Then out of the $6$ spots between the dogs, we have to put $5$ cats and this can be done in $$\binom{6}{5}$$ ways. Now in these $6$ ways, we will all sorts of combinations, one where say the $2$nd spot is not occupied, another where the $3$rd is left empty. As you do this you will realize why this is such a great method. It allows all possible combinations of boys together and apart while ensuring that the condition is met.