[Math] How many ways are there to place 10 distinct people within 3 distinct rooms with exactly 5 people in the first room and 2 people in the second room

combinatoricsdiscrete mathematicsinclusion-exclusion

So I was given this question. How many ways are there to place $10$ distinct people within $3$ distinct rooms with exactly $5$ people in the first room and $2$ people in the second room?

I have worked on a similar problem but it did not ask about the $5$ people in the first room and $2$ in the other, So that part is what is really throwing me off.

I started this off saying There are $3^{10}$ unique ways to assign $10$ distinct people in three rooms of which $3^{10} – 3 \cdot 2^{10} + 5$

After this I'm completely confused.

Best Answer

i believe the answer is ${10\choose 5}{5\choose 2}{3\choose 3}$ which can also be expressed as ${10\choose {5,2,3}}$.

$3^{10}$ would be if there were no rules as to how many go in each room