ThreeThree friends board an airliner just before departure time. There are only 1010 seats left, 44 of which are aisle seats. How many ways can the 33 people arrange themselves in available seats so that at least one of them sits on the aisle? The 3 people can arrange themselves in how many ways.

Na = 720 - 120 = 600 ways

The 3 people can arrange themselves in 600 possible ways.

Step-by-step explanation:

For at least one of the three friends to seat on an aisle seat, the number of possible arrangements can be given as;

Na = Nt - Nn 1

Where;

Na = number of possible arrangements with at least one of the seating on an aisle seat.

Nt = total number of possible arrangements for the three friends

Nn = number of possible arrangements with none of the three friends seating on an aisle seat.

Given;

Total number of seats available = 10

Total number of aisle seats available = 4

Total number of people = 3

So;

Nt = 10C3 = 10×9×8 = 720

Nn = (10-4)C3 = 6C3 = 6×5×4 = 120

From equation 1;

Na = 720 - 120 = 600 ways

The 3 people can arrange themselves in 600 possible ways.

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hope i !

p.s the first bit only matters if you know what that means.

p.p.s brainliest would really