If n objects labelled 1 to n are randomly placed in a row, what is the probability that exactly m of the objects will be correctly placed?

The equation is given below:

$\displaystyle c_{m,n} = {1 \over {m!}}[ {1 \over{2!}} - {1 \over{3!}} + {1 \over{4!}} .... +{ (-1)^{n-m} \over {n-m!}}]$

First, let's look at the case n=4 and m=4.

The equation correctly gives $\displaystyle {1 \over {4!}}$

Having all of the objects in their correct place is only 1 way out of 4! ways to arrange the 4 objects.

Can anyone explain to me the case when m=3 and n=4?

By right, it should return also $\displaystyle {1 \over {4!}}$, as if 3 out of 4 objects werein their correct place, then surely the last object must be in its place too.

But how does the formula make me see this?

Please can anyone help.