# Four people are boarding onto a mostly full airplane. There are 12 open seats, 2 of which are aisle seats. We want at least one of the four passengers to sit in an aisle seat. In how many ways can the 4 people sit?

6840 ways to have at least one person in an aisle seat (assuming each person and each seat is distinguishable).

#### Explanation:

We have 4 people looking to sit in some order in the remaining 12 seats of a full airplane. Two of those seats are aisle seats and we want at least 1 of the people to be sitting in an aisle seat.

I'm going to answer this from a permutation point of view - that is, we can distinguish each person and we can also distinguish each seat (aisle seat 31C is different than aisle seat 22D).

The general formula of a permutation is:

P_(n,k)=(n!)/((n-k)!); n="population", k="picks"

Let's first talk about the isle seats. We can either have one person in an aisle or two people in an aisle (and then calculate the rest of the possible seating after that). We need to look at each case separately:

2 aisle seat taken

I'm going to look at this case first because it's the more straightforward case.

We have 4 people looking to sit in 2 aisle seats. This gives us:

P_(4,2)=(4!)/((4-2)!)=(4!)/(2!)=24/2=12

The remaining 2 people are looking to sit in the remaining 10 seats:

P_(10,2)=(10!)/((10-2)!)=(10!)/(8!)=(10xx9xx8!)/(8!)=90

We can multiply these together and get:

$12 \times 90 = 1080$

1 aisle seat taken

For this case, we are looking to have one person out of the four sit in one of the two seats. This will give us:

P_(4,1)xxP_(2,1)=(4!)/(3!)xx(2!)/(1!)=24/6xx2/1=8

The remaining three people are looking to sit in the remaining 10 non-aisle seats:

P_(10,3)=(10!)/(7!)=(10xx9xx8xx7!)/(7!)=720

Multiply these together to get:

$8 \times 720 = 5760$

Putting it all together

We can now add the two cases together:

$1080 + 5760 = 6840$