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There are 3 passengers hoping to get a seat on the next flight to Chicago, but there are only 2 seats left on the plane. In how many ways can the seats be filled?

Question

There are 3 passengers hoping to get a seat on the next flight to Chicago, but there are only 2 seats left on the plane. In how many ways can the seats be filled?

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Solution

This is a problem of combinations. We are choosing 2 passengers (seats) from a group of 3 passengers. The formula for combinations is:

C(n, k) = n! / [k!(n-k)!]

where:

  • n is the total number of options,
  • k is the number of options to choose,
  • "!" denotes a factorial, meaning to multiply a series of descending natural numbers.

Here, n = 3 (the total number of passengers) and k = 2 (the number of available seats).

So, the calculation is:

C(3, 2) = 3! / [2!(3-2)!] = (321) / [(21)(1)] = 6 / 2 = 3

So, there are 3 different ways the 2 seats can be filled by the 3 passengers.

This problem has been solved

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