Counting/Permutation/Combination Help Please

[stagmite]

So I've been trying to teach myself probability and I've been working on all the problems in the book. However there are two problems in the counting/perm/combo chapter that I can't figure out.

The first problem I just need help getting started.

1. A row contains 12 chairs. In how many ways can 7 people be seated in these chairs.

For the second problem I thought this was a partition problem so I tried solving it by doing 9!/(3!3!3!), but that is absolutely wrong. So if someone can help me figure this out it'd be much appreciated.

2. A company has 9 analysts. Suppose the company divides the 9 analysts into 3 teams of 3 each, and each team works on the whole project. In how many ways can this be done?

Thanks for the help.

 

[soroban]

Hello, stagmite!

1. A row contains 12 chairs. In how many ways can 7 people be seated in these chairs.

Since the seven people are distintuishable, this is a permutation.

. . \(\displaystyle \begin{array}{c}\text{The 1st person has 12 choices of chairs.} \\ \text{The 2nd person has 11 choices of chairs.} \\ \text{The 3rd prson has 10 choices of chairs. } \\ \vdots \\ \text{The 7th person has 6 choices of chairs.} \end{array}\)


\(\displaystyle \text{There are: }\:12\cdot11\cdot10\cdot9\cdot8\cdot7\cdot6 \;=\;3,\!991,\!680 \text{ ways.}\)

2. A company has 9 analysts.
Suppose the company divides the 9 analysts into 3 teams of 3 analysts each,
and each team works on the whole project.
In how many ways can this be done?

Your approach was correct, but it needs one more step . . .


\(\displaystyle \text{The partition }\frac{9!}{3!\,3!\,3!} \:=\:1680\text{ applies if the three teams are }distinguishable.\)


\(\displaystyle \text{For example, the three teams might be: Research, Planning, and Implementation.}\)

. . \(\displaystyle \text{Then the partition }\{abc|de\!f|ghi\}\text{ is distinct from }\{de\!f|ghi|abc\}\)


\(\displaystyle \text{Assuming the three teams are indistinguishable (interchangable), we must divide by }3!\)

\(\displaystyle \text{Therefore, there are: }\:\frac{1680}{3!} \:=\:280\text{ ways.}\)

 

[stagmite]

Hello, stagmite!

1. A row contains 12 chairs. In how many ways can 7 people be seated in these chairs.

Since the seven people are distintuishable, this is a permutation.

. . \(\displaystyle \begin{array}{c}\text{The 1st person has 12 choices of chairs.} \\ \text{The 2nd person has 11 choices of chairs.} \\ \text{The 3rd prson has 10 choices of chairs. } \\ \vdots \\ \text{The 7th person has 6 choices of chairs.} \end{array}\)


\(\displaystyle \text{There are: }\:12\cdot11\cdot10\cdot9\cdot8\cdot7\cdot6 \;=\;3,\!991,\!680 \text{ ways.}\)



[quote:24fgq2m8]2. A company has 9 analysts.
Suppose the company divides the 9 analysts into 3 teams of 3 analysts each,
and each team works on the whole project.
In how many ways can this be done?

Your approach was correct, but it needs one more step . . .


\(\displaystyle \text{The partition }\frac{9!}{3!\,3!\,3!} \:=\:1680\text{ applies if the three teams are }distinguishable.\)


\(\displaystyle \text{For example, the three teams might be: Research, Planning, and Implementation.}\)

. . \(\displaystyle \text{Then the partition }\{abc|de\!f|ghi\}\text{ is distinct from }\{de\!f|ghi|abc\}\)


\(\displaystyle \text{Assuming the three teams are indistinguishable (interchangable), we must divide by }3!\)

\(\displaystyle \text{Therefore, there are: }\:\frac{1680}{3!} \:=\:280\text{ ways.}\)

[/quote:24fgq2m8]

Thank you so much soroban. I've been racking my brain over these problems since yesterday, but now that you explained it I finally understand where I went wrong. Hopefully I'll recognize these sort of problems better in the future. BTW is there any good online probability Q/A that I can practice on? I'm actually trying to teach myself probability because I want to change careers and do something in probability and statistics.

 

[soroban]

To all my critics (including a few Mods):

Thank you so much soroban. I've been racking my brain over these problems since yesterday,
but now that you explained it I finally understand where I went wrong.
Hopefully I'll recognize these sort of problems better in the future.

THIS is the response I hope for when I give full solutions.

I see no progress in constantly scolding, "Show us some work or we won't help you!"
. . or "We don't do your homework here!"

I could do that . . . makes for less work for me.
99% of them don't respond and go to a friendlier site.

Is that the desired image for FreeMathHelp?
Like so many clerks today (sorry, I mean "retail associates"),
. . instead of "May I help you?", it's more like, "Yeah, whaddaya want?"
(They aren't there to serve . . . you're there to ruin their free time.)

You might ask, "Do you have one of those, y'know, things that go in the air filter and ..."
Would you appreciate: "Look, when you get your question organized, come see me.
. . Show some effort. .Right now, you're wasting my time and annoying me."

Think about it.

If you do, maybe I'll change my signature.

 

[galactus]

That, too, is my stance. Recently, I gave a student a full solution to a Newton's cooling problem. Now, that they have a template they can get the idea

how it works and solve future prolems like that. I will not keep doing them one after another, but I like to give one nice example so they have something to

show them how to do a problem. Giving vagueries and small hints ususally succeeds in having a dozen posts on one topic when the poster does not understand.

That is the way I am going to keep doing it. No, I am not going to get into a debate about this. Just expressing my opinion.