Solving Permutation Question: 6 Men, 3 Women

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In summary, there are 9! ways to arrange 6 men and 3 women without any restrictions. To arrange them with no 2 women next to each other, you can first arrange the men in 6! ways and then distribute the 3 women in 7P3 ways, giving a total of 6! * 7P3 ways.
  • #1
rock.freak667
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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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  • #2
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?
 
  • #3
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 ?
 

1. How many different arrangements can be made with 6 men and 3 women?

There are 9 people in total and we need to arrange them, so the number of arrangements is 9!. This can also be written as 9 factorial which is equal to 362,880.

2. What is the probability of selecting 2 men and 1 woman in a random arrangement?

The total number of ways to select 2 men and 1 woman from a group of 6 men and 3 women is 6C2 * 3C1. This can also be written as 6 choose 2 for the men and 3 choose 1 for the woman. The total number of ways is 15 * 3 = 45. The probability is then 45/362,880 which simplifies to 1/8,064 or approximately 0.0124 or 1.24%.

3. How many different arrangements are possible if the men and women must be seated alternately?

In this case, we have 6 men and 3 women, so we can think of them as one group of 6 and another group of 3. The number of arrangements for each group is 6! and 3!. However, since they must be seated alternately, we need to multiply by 2 because there are two options for the starting group (men or women). Therefore, the total number of arrangements is 2 * 6! * 3! = 1,814,400.

4. How many different arrangements are possible if 2 specific men must sit together?

If we consider the 2 specific men as one person, we now have 8 people to arrange. The number of arrangements is then 8!. However, the 2 specific men can also be arranged within their pairing, so we need to multiply by 2. Therefore, the total number of arrangements is 2 * 8! = 806,400.

5. What is the probability of selecting 3 men and 2 women in a random arrangement if the women must sit together?

We can first calculate the total number of ways to select 3 men and 2 women from a group of 6 men and 3 women, which is 6C3 * 3C2 = 20 * 3 = 60. However, since the women must sit together, we can consider them as one person. So, the total number of ways to select 3 men and the pair of women is 6C3 * 2C1 = 20 * 2 = 40. The probability is then 40/362,880 which simplifies to 1/9,072 or approximately 0.011%.

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