Solving Permutation Question: 6 Men, 3 Women

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To solve the permutation problem of arranging 6 men and 3 women, the total arrangements without restrictions is calculated as 9!, equating to 362,880 ways. For the arrangement where no two women are adjacent, the approach involves first arranging the 6 men, which can be done in 6! ways. This creates 7 potential gaps for placing the women, and the task is to choose 3 out of these 7 gaps to place the women, calculated as 7P3. The discussion highlights the complexity of ensuring no two women are next to each other while considering overlapping cases where women might cluster together. Overall, the solution requires careful combinatorial reasoning to avoid double counting.
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Homework Statement


6 men and 3 women are arranged in a line
Find the number of ways:
A)That they can be arranged without any restrictions
B)They can line up with no 2 women next to each other


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The Attempt at a Solution


A)Well that is simply 9!
B)This is where it is hard...I thought to find the number of ways that 3 women next to each other could have been arranged and then subtract this from 9!. But then I also have to subtract with 2 women next to each other,but there is a problem if I find this,this accounts for that there could be an instant when there are 3 women next to each other.


any help on this part?
 
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For b), I think I'd do it as follows:
First choose an order for the men, this gives you 6! possibilities:
. m . m . m . m . m . m .
On the dots you can place a woman (but no two, then you would have two next to each other). How many ways are there to distribute 3 women over the seven empty spots?
 
Well then you can do that 7P3*6!

but then wouldn't that be very long to write out all of the combinations of M ?
 
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