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Ne4ueva [31]
3 years ago
9

How many ways are there for eight men and five women to stand in a line so that no two women stand next to each other

Mathematics
1 answer:
Sloan [31]3 years ago
4 0

Answer:

609638400 ways

Step-by-step explanation:

For 8 men in a line: 8! = 40320

<u>   </u>O<u>   </u>O<u>   </u>O<u>   </u>O<u>   </u>O<u>   </u>O<u>   </u>O<u>   </u>O<u>   </u>

1     2    3    4   5    6   7     8   9 ... in order to separate 2 women, there are 9 vacant places for 5 women, we have to choose 5 out of 9

C(9,5) = 9! / 5!4! = 126

Random place 5 women in the 5 chosen places: 5! = 120

Total ways: 40320 * 126 * 120 = 609638400

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