How many ways to distribute n people in r lines

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galactus

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I seen a probability problem on another site I thought was interesting.

"How many ways can 6 people line up at 2 cash registers".

The response was 3912 ways, but is this correct?.

Order should matter. I count 3*1440+720=5040 ways.


There should be a general way to do this. The number of ways to distribute n people in r lines. I would think this would be kind of like distributing n distinct objects into r distinct boxes where order matters.
 
Look at the given answer in factored form: \(\displaystyle 3912 = 2^3 \cdot 3 \cdot 163.\)
That makes no sense as a counting formula.

I find the question to be a bit ambiguous. As written, I agree with you that the answer is 5040, that is (7!). See we can ‘break’ any line of six in seven places. In this model some register my by unused.

If you use a bank teller model, line up and take the next available teller, then both of the two tellers would be used. That can be done in 5(6!)=3600 ways.

Maybe there is more to the problem?
 
Hey pka, I was correct then, depending on how the problem is taken.

I didn't think the posted response was correct, but....

Here's the link. Take a peek. :D

It just says, "how many ways can 6 people line up to pay at two different cashiers?".

http://math2.org/mmb/thread/38633
 
That is an undercount. I see what he is doing.
Note that \(\displaystyle \sum\limits_{j = 0}^6 {\left( {_6 P_j } \right)(6 - j)!} = 5040.\)
 
Wow, I counted right then. A blind hog finds an acorn once in awhile :D

I didn't use the formula you posted(which I will remember). I counted up each scenario.

I placed 6 people at one register and arranged them in 6! ways. Times 2 because of the 2 registers. That is 1440.

Then, I put 5 at a register and 1 at another. That's 5!*6*2=1440

Then, 4 people at a register and 2 at another. That's P(6,2)*4!*2=1440

Then, 3 at a register and 3 at another. That's P(6,3)*3!=720

This totals 5040.
 
galactus said:
The response was 3912 ways, but is this correct?.
Click to expand...

Isn't "the response":
2*6P6 +2*6P5 +2*6P4 +2*6P3+ 2*6P2 +2*6P1
equal to 126 ?
 
Yes the response is \(\displaystyle \sum\limits_{j = 1}^6 {(2)\left( {_6 P_j } \right)} = 3912.\)
But that is an undercount. See the reasons above.
 
No, I get 3912

P(6,6)=720
P(6,5)=720
P(6,4)=360
P(6,3)=120
P(6,2)=30
P(6,1)=6

2(720+720+360+120+30+6)=3912
 
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