How Many Ways Can 11 People Be Seated at Two Round Tables?

[member 428835]

Homework Statement

11 people sit at two round tables, one sits 5 and the other sits 6. how many combinations of seating arrangements are there?

Homework Equations

i'm not sure about equations, but my solution attempt may have some info.

The Attempt at a Solution

i know if 11 people were to sit at 1 table, we would have \frac{11!}{11}=10! combinations. thus, would we first have to choose who sits at which table? i.e: (11C5)(4!)+(11C6)(6!)

thanks for your time (and help!)

 

[haruspex]

You're close, but a couple of errors.
You got ¹¹C₅4! for choosing the 5 to sit at one table and arranging them. For each such arrangement, how can you arrange the remaining 6?

 

[member 428835]

wait, do you mean (11C5)(4!)+(11C6)(5!)? also, if we've already selected the 5 to be at on table, do we really need to select 6 for the other?

thanks for the response, and let me know what you think.

 

[member 428835]

because i definitely should have done 5!, not 6! (brain fart)

 

[D H]

There are two big problems with your result. One is that when you've chosen the five people who will sit at table #1 you have but one choice as to which set of six people will sit at table #2. The other: Why are you adding?

 

[member 428835]

There are two big problems with your result. One is that when you've chosen the five people who will sit at table #1 you have but one choice as to which set of six people will sit at table #2.

yea, i definitely see this problem but if i don't put both "choices" how do i determine which number (5 or 6) to use?

also, i was adding because i figured after i choose who sits at what table, all i need to do is take the possible combos of one table and add them to the combos of another table.

evidently this is wrong; can you direct me as to what to do next?

thanks for your help!

 

[D H]

yea, i definitely see this problem but if i don't put both "choices" how do i determine which number (5 or 6) to use?

There are a number of right ways to address this part of the problem. You chose a wrong way.

One way to look at this part of the problem is that you choose five people to sit at one table, call it table A. There are 11 choose 5 ways to do this. For any give choice, there are 11-5=6 people left. You now need to choose six of these six people to sit at table B. There are 6 choose 6 ways to do this, and that of course is one. The number of ways to choose people to sit at the two tables is the product of these two numbers: (11 choose 5)*(6 choose 6).

You could of course look at it the other way around: You'll choose six people to sit at the larger table and then choose five of the remaining five to sit at the smaller table. This leads to (11 choose 6)*(5 choose 5) ways to split the people into two groups.

Hmmm. One approach yields (11 choose 5)*(6 choose 6), the other (11 choose 6)*(5 choose 5). Is there a problem here? The answer is no. 11 choose 5 and 11 choose 6 are the same number (462), as are 6 choose 6 and 5 choose 5 (which are of course both 1).

also, i was adding because i figured after i choose who sits at what table, all i need to do is take the possible combos of one table and add them to the combos of another table.

Suppose X and Y are independent. Also suppose there are three ways to accomplish X and five ways to accomplish Y. The number of ways to do accomplish X and Y is fifteen, not eight.

The same applies to the problem at hand. You should be multiplying here, not adding.

 

[CAF123]

yea, i definitely see this problem but if i don't put both "choices" how do i determine which number (5 or 6) to use?

You can do it either way (i.e choose 6 for one table and have one set of five for the other, or choose 5 for a table and have one set of six for the other). The result is the same.

also, i was adding because i figured after i choose who sits at what table, all i need to do is take the possible combos of one table and add them to the combos of another table.

Consider the case when there are two people situated at one table and one at the other. How many possible combinations are there?

Alternatively, are the combinations around each individual table dependent on each other? What does this tell you?