Inviting Friends Over Combinations/Permuations Assume we have three friends and we want to invite each to dinner for 6 days. I want to invite one for 3 days, another for 2 days and another for 1 day. The answer is $P(3,3) \times \binom{6}{3} \times \binom{3}{1} $ I understand that there are three ways to decide which person to invite for three nights, and then we want to choose a person to invite for 3 nights out of 6 and another person to invite for one night out of three. I kind of don't like this approach of "leaving out" a possibility. So first, how does the above equation take into the account the `two remaining days` and what would be a different way to `account for` those two days in the equation?

There 3 three people, 1 takes 3 days, 1 takes 2 days, 1 takes 1 day.

We can understand the above answer as :

First you can permute 3 people's days after you choose their schedule.

As you said, we choose 3 out of 6 days for the first person. We remain 3 days.

Then we choose 1 out of 3 days for the third person.

Then we choose the last two days for the second person. The answer is:

3! * C(6,3) * C(3,1) * C(2,2)

C(2,2) = 1, and they did not put it in the answer.