Permutation: $5$ into $8$ if only $2$ can share

Question

I am having trouble figuring out a permutation problem:

"In how many ways can $5$ mathematicians be put into $8$ offices, where each mathematician has an office to themselves? What if only $2$ of the mathematicians cannot share an office with anyone?"

So I'm thinking for the first part $n=8$ and $k=5$. So the number of ways would be $8\times7\times6\times5\times4 = 6720$

But how would you figure out how many permutations there are if $2$ are allowed to share but the rest cannot?

They should be: it's like multicolored balls and numbered bins — Alex, Oct 14 '15 at 12:54
You have contradicting statements. First statement: "only 2 of the mathematicians cannot share an office with anyone". Second statement: "2 are allowed to share but the rest cannot". — barak manos, Oct 14 '15 at 14:10

score 0 · Accepted Answer · answered Oct 14 '15 at 13:23

The number of permutations includes those where the two do not share with each other ($6720$ as you had), as well as those where they do (let this be $N_s$).

To compute $N_s$ note that this is equivalent to placing four mathematicians into eight offices, and noting that the position of the fifth is determined by their partner's office.

$$N_s=8\times7\times6\times5=1680$$

Hence the total is $6720+1680=8400$.

Permutation: $5$ into $8$ if only $2$ can share

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