How many ways are there for four men and five women ...

[r0bHadz]

Homework Statement

How many ways are there for four men and five women to stand in a line so that

All men stand together?

All women stand together?

Homework Equations

The Attempt at a Solution

For all men stand together, you can group the 4 men as one token, then there are (1+5)! ways the men and women can stand in a line, but the four men can be arranged in 4! ways so the answer would be (1+5)!(4!) = (6!)(4!)

similar for women
let 5 women = one token, then you have (1+4)! but the 5 women can be arranged 5! ways so you have (5!)(5!) as the answer

is there anything I am missing?

 

[PeterDonis]

Moderator's Note: Thread moved to pre-calculus math homework forum.

 

[WWGD]

Men can go in positions 1-5, 2-6, etc.

 

[r0bHadz]

Men can go in positions 1-5, 2-6, etc.

So I am taking the 4 men as one line. There are 4! possible combinations for the men. Then, There are 5 women, and the group of four men which I'm considering as 1. There are 6! possible arrangements here.

Wouldn't "men can go in positions 1-5,2-6, etc" be under the (6!)?

 

[PeroK]

Homework Statement

How many ways are there for four men and five women to stand in a line so that

All men stand together?

All women stand together?

Homework Equations

The Attempt at a Solution

For all men stand together, you can group the 4 men as one token, then there are (1+5)! ways the men and women can stand in a line, but the four men can be arranged in 4! ways so the answer would be (1+5)!(4!) = (6!)(4!)

similar for women
let 5 women = one token, then you have (1+4)! but the 5 women can be arranged 5! ways so you have (5!)(5!) as the answer

is there anything I am missing?

That's right. You could have tested your approach with smaller numbers, perhaps 2 and 3.

 

[r0bHadz]

That's right. You could have tested your approach with smaller numbers, perhaps 2 and 3.

Sorry are both answers correct? WWGD's post has me a little paranoid lol

 

[PeroK]

Sorry are both answers correct? WWGD's post has me a little paranoid lol

Yes, the answers are correct.

That's another way to do it. For the men as a group, they have 6 possible positions (with the first man in position 1-2-3-4-5 or 6). Then it's $6 \times 4! \times 5!$, which is the same as you got.

 

[WWGD]

Sorry are both answers correct? WWGD's post has me a little paranoid lol

Don't worry, my reply agrees with PeroK's and yours.