A blue plane, a red plane and a white plane are waiting to take off from an airport that has two runways. Planes must take off from the airport one at a time, but can take off from either runway. In how many ways can the three takeoffs be scheduled? (One such way is the blue plane on runway A, followed by the red plane on runway B, followed by the white plane on runway B.)

in 48 ways.

Step-by-step explanation:

So we have 3 takeoffs.

3 planes, and 2 runways

For the first takeoff, we have 3 options to choose (red plane, blue plane and white plane) And for the runway, we have 2 options (A or B)

So we have 3*2 = 6 options

For the second takeoff, we have 2 options to choose (because one of the planes was already selected before) and we still have 2 options for the runway.

So we have 2*2 = 4 options

For the third takeoff, we have 1 plane to choose from, and for the runway, we have 2 options.

So we have 1*2 = 2 options

The total number of combinations is equal to the product of the number of options in each selection:

C = 6*4*2 = 48 combinations

except a rational number. : )

what does x eaqual

step-by-step explanation: