Total number of methods to arrange 7 girls and 5 boys in a row in such a way that no two girls can sit together ?
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First, seat have the 7 boys sit down in a row of 7 seats. There are 7!=5040 such arrangements.
Next, put a seat between each boy, and to either side of the boys on either end of the row - there should be 8 such empty seats.
Choose 5 of the 8 seats for the girls to sit in: (85)=56. Then arrange the 5 girls in the 5 seats. There are 5!=120 ways to do so. 56∗120=6,720. Alternately, you could just have each girl choose one of the 8 seats. So the first girl would have 8 choices, the second would have 7 since the first girl is sitting in one of the 8, the third would have 6 choices, the fourth 5, the fifth 4. So you have 8∗7∗6∗5∗4=6,720.
Finally, combine it all: 5040∗6720=33,868,800 total possible arrangements.
hannjr:
The boys can be arranged in 7! ways = 5040 and the girls can be arranged in 5! = 120 ways giving 5040 * 120 = 604,800 ways for a particular seating arrangement. Take the arrangement: BBGBGBGBGBGB and move the boy on the right one position giving BGBBGBGBGBGB and again move the boy on the right one position giving BGBGBBGBGBGB. You can do this 6 times giving 6 * 604,800 = 3,628,800 arrangements.
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