Distinguishable Permutations (14 construction workers with 7 for mixing cement, 5 for laying bricks, 2 for carrying bricks)

[jpanknin]

I'm using Stewart's Algebra and Trig, 4E. In a section on distinguishable permutations, the book gives two examples. One is 15 balls, 4 red, 3 yellow, 6 black, 2 blue. The answer is: [math]\frac{15!}{4!*3!*6!*2!} = 6,306,300[/math]
The next example is 14 construction workers with 7 for mixing cement, 5 for laying bricks, 2 for carrying bricks. The answer given in the book is [math]\frac{14!}{7!*5!*2!}=72,072[/math]
I'm fine with the first example because a red ball can't also be a yellow ball. But for the workers it seems that there would need to be some kind of constraint on which group the workers were placed in (had to be certified to mix cement, for example) for this permutation to be "distinguishable." But if all 14 workers were qualified to do all three jobs, wouldn't the answer just be 14!? Just because they're assigned different roles doesn't necessarily mean that they're "distinguishable," does it? Or am I missing something?

 

[Steven G]

You are talking about answers in the back of the book, yet you have not stated the question

Seriously, if someone tells you 15 balls, 4 red, 3 yellow, 6 black, 2 blue you would understand the question. I hope not as no question was asked.

I claim that the answer is 4/30. Why, because 4 out of the 30 balls are blue.

 

[Steven G]

I'm fine with the first example because a red ball can't also be a yellow ball. But for the workers it seems that there would need to be some kind of constraint on which group the workers were placed in (had to be certified to mix cement, for example) for this permutation to be "distinguishable." But if all 14 workers were qualified to do all three jobs, wouldn't the answer just be 14!? Just because they're assigned different roles doesn't necessarily mean that they're "distinguishable," does it? Or am I missing something?

The 14 workers have been defined as 7 for mixing cement, 5 for laying bricks, 2 for carrying bricks.
The 7 cement mixers for example can't be just come from any of the 14 worker.
It is exactly the same as the 1st problem. Just use 7 red, 5 blue and 2 yellow instead.

 

[jpanknin]

You are talking about answers in the back of the book, yet you have not stated the question

Seriously, if someone tells you 15 balls, 4 red, 3 yellow, 6 black, 2 blue you would understand the question. I hope not as no question was asked.

I claim that the answer is 4/30. Why, because 4 out of the 30 balls are blue.

I'm confused. These are examples from the chapter, not the back of the book. The exact questions are:

Example 7: Find the number of ways of placing 15 balls in a row given that 4 are red, 3 are yellow, 6 are black, and 2 are blue.

Example 8: Fourteen construction workers are to be assigned to three different tasks. Seven workers are needed for mixing cement, five for laying bricks, and two for carrying the bricks to the brick layers. In how many different ways can the workers be assigned to these tasks?

The solutions to each example are then given exactly as shown in my original post.

 

[pka]

I'm confused. These are examples from the chapter, not the back of the book. The exact questions are:
Example 7: Find the number of ways of placing 15 balls in a row given that 4 are red, 3 are yellow, 6 are black, and 2 are blue.
Example 8: Fourteen construction workers are to be assigned to three different tasks. Seven workers are needed for mixing cement, five for laying bricks, and two for carrying the bricks to the brick layers. In how many different ways can the workers be assigned to these tasks?
The solutions to each example are then given exactly as shown in my original post.

Say we use the codes [imath]{\bf M}\text{ for mixing, }{\bf L}\text{ for laying, and}{\bf C}\text{ for carring}.[/imath]
Then make a list of the names of the fourteen workers Using the string [imath]{\bf MMMMMMMLLLLLCC}[/imath]
along side the names to make the assignments. That string can be arranged in [imath]\dfrac{14!}{(7!)(5!)(2!)}[/imath] ways.

[imath][/imath]

 

[mmm4444bot]

I'm confused … The exact questions are … Find the number of ways of placing 15 balls in a row given that 4 are red, 3 are yellow, 6 are black, and 2 are blue.

Example 8: Fourteen construction workers are to be assigned to three different tasks. Seven workers are needed for mixing cement, five for laying bricks, and two for carrying the bricks to the brick layers. In how many different ways can the workers be assigned to these tasks?

Hello. The confusion arose because you'd provided the blue information only, in post#1. You had omitted the red information (i.e., the "question" parts).

Please follow the posting guidelines. Thank you!

Post the complete text of the exercise. This would include the full statement of the exercise…

[imath]\;[/imath]

 

[Steven G]

I'm confused. These are examples from the chapter, not the back of the book. The exact questions are:

Example 7: Find the number of ways of placing 15 balls in a row given that 4 are red, 3 are yellow, 6 are black, and 2 are blue.

Example 8: Fourteen construction workers are to be assigned to three different tasks. Seven workers are needed for mixing cement, five for laying bricks, and two for carrying the bricks to the brick layers. In how many different ways can the workers be assigned to these tasks?

The solutions to each example are then given exactly as shown in my original post.

Yes, you need to post the exact problem. Why would anyone assume that you want to know in how many ways can you place 15 balls in a row if you have 3 red, 3 yellow, 6 black and 2 blue if all you state is you have 3 red, 3 yellow, 6 black and 2 blue.

A very fair question could have been how many balls are not blue or how many more black balls than yellow balls.
You need to ask a question--which you did in post 4

As far as you being confused what I was saying is how can there be an answer in the back of the book if no question was asked?

 

[jpanknin]

Yes, you need to post the exact problem. Why would anyone assume that you want to know in how many ways can you place 15 balls in a row if you have 3 red, 3 yellow, 6 black and 2 blue if all you state is you have 3 red, 3 yellow, 6 black and 2 blue.

A very fair question could have been how many balls are not blue or how many more black balls than yellow balls.
You need to ask a question--which you did in post 4

As far as you being confused what I was saying is how can there be an answer in the back of the book if no question was asked?

Got it and understood. Will do so in the future.

 

[jpanknin]

Say we use the codes [imath]{\bf M}\text{ for mixing, }{\bf L}\text{ for laying, and}{\bf C}\text{ for carring}.[/imath]
Then make a list of the names of the fourteen workers Using the string [imath]{\bf MMMMMMMLLLLLCC}[/imath]
along side the names to make the assignments. That string can be arranged in [imath]\dfrac{14!}{(7!)(5!)(2!)}[/imath] ways.

[imath][/imath]

The way I was thinking about it is if, out of the 14 workers, only 7 are qualified (or some constraint on which workers can do which job) can mix cement, then the 7! makes sense. But if all 14 workers could technically do all 3 jobs, then for the first role in your list above "M," 14 workers would be available to fill that slot. I understand that only 7 will be assigned to mix cement, but if all 14 COULD mix cement, then they have 14 workers to initially choose from if the first M was the first role to be assigned.

 

[Steven G]

You have the 14 people lined up.
In front of each person you put an M, L or C (exactly 7 M's, exactly 5 L's and exactly 2 C's)
One such string will me MCMLMLMLCLMLMM. So the problems becomes in how many ways can you arrange 7M's, 5L's and 2 C's in a line.
This is exactly what pka posted.

The problem stated that 14 workers were to be assigned to ....
Nobody said that anyone of the workers would be able to do the job that they were assigned to. That is where you are confused and I can understand why.

 

[Steven G]

But if all 14 workers could technically do all 3 jobs, then for the first role in your list above "M," 14 workers would be available to fill that slot. I understand that only 7 will be assigned to mix cement, but if all 14 COULD mix cement, then they have 14 workers to initially choose from if the first M was the first role to be assigned.

That is 100% correct.

 

[jpanknin]

You have the 14 people lined up.
In front of each person you put an M, L or C (exactly 7 M's, exactly 5 L's and exactly 2 C's)
One such string will me MCMLMLMLCLMLMM. So the problems becomes in how many ways can you arrange 7M's, 5L's and 2 C's in a line.
This is exactly what pka posted.

The problem stated that 14 workers were to be assigned to ....
Nobody said that anyone of the workers would be able to do the job that they were assigned to. That is where you are confused and I can understand why.

Got it. That makes more sense. Thank you.

 

[NotJeffM]

Please do not allow your knowledge of the real world intrude into your analysis of math word problems.

 

[ImMAD]

I think the question in the book is poorly phrased, because in contrast to the colourful balls question, it does not say that the workers need or are to be lined up. So being confused is the right answer.

 

[Dr.Peterson]

I think the question in the book is poorly phrased, because in contrast to the colourful balls question, it does not say that the workers need or are to be lined up. So being confused is the right answer.

There need not be any mention of "lining up". Jobs are being assigned to people, so it is equivalent to numbering the people and assigning a set of numbers to each job, or putting a set of letters representing the jobs into a row of numbered positions:

MMMMMMMLLLLLCC --> _ _ _ _ _ _ _ _ _ _ _ _ _ _

Example 8: Fourteen construction workers are to be assigned to three different tasks. Seven workers are needed for mixing cement, five for laying bricks, and two for carrying the bricks to the brick layers. In how many different ways can the workers be assigned to these tasks?

In probability and combinatorics, it is very common to find an equivalent problem like this. Here, imagining the people being lined up is a useful tool (not part of the problem statement, but of our thinking).

 

[ImMAD]

There need not be any mention of "lining up". Jobs are being assigned to people, so it is equivalent to numbering the people and assigning a set of numbers to each job, or putting a set of letters representing the jobs into a row of numbered positions:

MMMMMMMLLLLLCC --> _ _ _ _ _ _ _ _ _ _ _ _ _ _



In probability and combinatorics, it is very common to find an equivalent problem like this. Here, imagining the people being lined up is a useful tool (not part of the problem statement, but of our thinking).

You are right, that makes sense.