Solving Permutation Question: 6 Men, 3 Women

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SUMMARY

The discussion focuses on solving a permutation problem involving 6 men and 3 women arranged in a line. For part A, the total arrangements without restrictions is calculated as 9!, equating to 362,880 ways. In part B, the challenge is to arrange the individuals such that no two women are adjacent. The proposed solution involves arranging the men first (6!) and then placing the women in the available gaps, leading to the calculation of 7P3 * 6! for the valid arrangements.

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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


Homework Equations





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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