Question 1199761: i) In how many ways six men and four women sit in a row
ii) In how many ways can they sit in a row if all the men sit together
iii) In how many ways can they sit in a row if just the women sit together
iv) In how many ways can they sit in a row if men sit together
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
Part (i)
6 men + 4 women = 10 people total
10 ways to fill the first slot
9 to fill the second slot
8 to fill the third slot
and so on
10*9*8*7*6*5*4*3*2*1 = 10! = 3,628,800
The exclamation mark indicates factorial.
Answer: 3,628,800
This value is roughly 3.63 million.
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Part (ii)
Let's say the four women have code names A,B,C,D
We'll have the men temporarily leave the group.
Another woman with code name "E" will be introduced.
Wherever woman E is located, she represents the block of all men sitting together.
We have five letters A,B,C,D,E
with 5*4*3*2*1 = 5! = 120 permutations
Within any of these 120 permutations, there are 6! = 720 ways to rearrange the men.
Therefore, we have 5!*6! = 120*720 = 86,400 different ways to arrange the people such that the men stay together.
Answer: 86,400
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Part (iii)
We'll follow very similar logic as compared to the previous part.
Let's say the six men have code names A,B,C,D,E,F
The women leave the group temporarily, and in their place a man with code name "G" steps in.
Wherever man G is, the block of women will step in his place.
There are 7 letters in A,B,C,D,E,F,G with 7! = 5040 permutations.
Within each of those permutations, there are 4! = 24 ways to arrange the women in their block.
7!*4! = 5040*24 = 120,960 total ways to arrange the people if all the women stay together. This time the men can separate.
Answer: 120,960
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Part (iv)
This part seems like a repeat of part (ii).
Please double-check to make sure you typed the question correctly.
If there isn't a typo, then this question has very similar wording to part (ii).
The key omission here is the word "all".
I interpret this to mean that some of the men can sit together, while others do not. Perhaps we have 2 or more blocks of men. Or we have 1 pairing of men, and the rest of the men are separate islands so to speak.
M = man
W = woman
A sequence like M,W,M,M has one man on his own island, while the last two men are grouped together.
The two sets of men are separated by a woman in the second slot.
Let's try to construct a scenario in which all of the men are separated. In other words: we'll try to form 6 separate islands of men.
There are 4 women. This gives the sequence
_, W, _, W, _, W, _, W, _
The blanks represent a potential spot for a man.
We have 4 women and 4+1 = blank spots.
We update to
M, W, M, W, M, W, M, W, M
The 6th man will go somewhere in this sequence.
No matter how hard we try, there's no way to have this 6th man get his own island. He'll be grouped with one of the initial 5 men already in the sequence.
What to take away from this: It's impossible to have the 6 men be fully separated. Meaning that each of the 10! = 3,628,800 different permutations will involve at least a pairing of two men.
Once again, the wording for part (iv) is similar to the wording in part (ii); which makes me think there might be a typo.
Or the wording seems a bit vague.
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