How many ways can I arrange $x$ zeroes in $n$ spaces

Question

I cant figure out the formula for arranging $x$ zeroes in $n$ spaces as shown below

for 3 spaces and 2 zeroes

0 0 -
0 - 0
- 0 0


for 4 spaces 2 zeroes
0 0 - -
- 0 0 -
- - 0 0
0 - 0 -
- 0 - 0
0 - - 0

Any help would be appreciated

Answer 1

Okay.....

Method 1: We have $x$ zeros.

How many spots are there to put the first $\color{blue}0$ which for notation purposes I will color blue. There are $n$ space.

How many spots are there to put the second $\color{green}0$ which for notation purposes I will color green. We already filled a spot. So there are $n-1$ spaces available.

How many spots are the to put the third $\color{red}0$. We already filled two so there are $n-2$ spaces available.

And so on.

So in the end there are $n*(n-1)*(n-2)*...... *(n-(x-1))$ ways to arrange these $x$ different colored $0$.

And while we are at it... $n*(n-1)*(n-2)*...... *(n-(x-1)) =\frac {n*(n-1) *.......*(n-(x-1))*(n-x)* ...... *2*1}{(n-x)*.....*2*1} =\frac{n!}{(n-x)!}$.

So let's look at one of our results.

$---\color{blue}0-\color{red}0\color{purple}0--\color{green}0-$.

We first chose the 4th spot for the blue $0$ and then we chose the $10$th spot for the green $0$ and then we chose the tthe 6th spot for the red $0$.

Well that's exactly the same thing as if you had chosen

$---\color{purple}0-\color{green}0\color{red}0--\color{blue}0-$

where we chose the exact same spots but in a different order.

How many ways are there to choose the same $x$ spots but in a different order.

Well, there were $x$ zeros we could have chosen for the first spot And $x-1$ for the second and so on until we have just one spot left for the last zero.

So there are $x*(x-1) *....*2*1 = x!$ ways to arrange the zeros in the $x$ spots.

So as all of those arrangements for those particular spots are considered to actually be the exact same thing, we must divide our result of $n*..... *(n-(x-1))!$ but $x!$ as the $n*..... *(n-(x-1))!$ results will contain $x!$ different occurances of what are actually the same thing if we ignore color (which only existed for notation purposes).

So there $\frac {n*..... *(n-(x-1))!}{x!} = \frac {n!}{x!(n-x)!}$.

We call that number ${n\choose x}$ and we call it "$n$ choose $x$" which literally means the number of ways we can choose $x$ positions from $n$ slots (or $x$ balls from a bag of $n$ balls....whatever.