# In a waiting room are 10 seats. Three sisters want to sit together and there are three other people as well. In how many ways can the 6 people sit in the waiting room?

10,080 ways

#### Explanation:

Before we move forward with this question, I'm going to assume that the seats are all in a row (although what waiting room has 10 seats in a row is a question for another day...). I'm also going to assume that each person is an individual and so having one sister or another sister sit in a seat makes a difference (vs simply having a sister sit in a seat).

Let's do this with a bit of a visual. First off we have 10 seats:

$1 \textcolor{w h i t e}{0} 2 \textcolor{w h i t e}{0} 3 \textcolor{w h i t e}{0} 4 \textcolor{w h i t e}{0} 5 \textcolor{w h i t e}{0} 6 \textcolor{w h i t e}{0} 7 \textcolor{w h i t e}{0} 8 \textcolor{w h i t e}{0} 9 \textcolor{w h i t e}{0} 10$

Ok - now first let's talk about the 3 sisters: (A)bby, (B)arb, (C)arol. They want to sit together - so that means they can sit in seats: (1,2,3), (2,3,4), (3,4,5), etc - in all, there are 8 groups of seats they can take up.

Within that group of seats, the sisters can sit (3!)=6 ways, so the sisters can sit together $8 \times 6 = 48$ ways.

Now we have 7 seats left and 3 people to fill them. Again, we'll look at this as a Permutation:

P_(7,3)=(7!)/((7-3)!)=(7!)/(4!)=(7xx6xx5xx4!)/(4!)=210 ways for each of the 48 ways the sisters can be seated.

And so there are $48 \times 210 = 10 , 080$ ways they can be seated.