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In how many ways can 4 girls and 2 boys sit at a movie theater row with 6 seats if a girl must be seated at each end.

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Step 1: Pick a girl and seat her at the left end, 4 ways.

Step 2: Pick a girl from the remaining 3 and set her at the right end, 3 ways.

Step 3: Four people remain; assign them to the remaining 4 seats: 4! ways.

Now multiply and get 12*24 = 288 possible seatings.

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thanks for the help also what does 5!.3! mean in permutations. – Adam Jul 8 '12 at 14:48

If a girl must be at either end, then the only seat options for the boys are the middle 4 seats. The number of ways of assigning two of the middle 4 seats to the 2 boys is $${}_4P_2=4\times 3=12.$$

Now we have to choose where the girls sit. The number of ways of assigning the remaining 4 seats to the 4 girls is $${}_4P_4=4\times 3\times 2\times 1= 24.$$

Finally, multiply these results to get $12\times 24=288$ different seating arrangements where girls are on either end.

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