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There are 3 tables that can each sit 5 people: one facing north, one facing south, and one facing east; 10 people are to be seated at random. a) How many different arrangements are possible to sit the 10 people? b) How many arrangements are possible if Joe and Frank don't want to sit next to each other? c) How many arrangements are possible if Joe and Frank don't want to sit next to each other or across from each other?

My attempts: a) I thought this would be the same as the number of ways of putting 10 distinct balls in 15 numbered boxes, with at most 1 ball in each box. Because we are interested in the final combinations, I thought the answer would be ${15}\choose{10}$

b) Let $A_{edge}$ be the even that Joe/Frank sit next to each other at the edges of one of the 3 tables: choose a table, choose an edge (right or left), and choose if Joe or Frank sits at edge. Then arrange remaining 8 people in available 13 seats; $|A_{edge}|$=3*2*2*${13}\choose{8}$ Let $|A_{mid}|$ be that Joe/Frank sit next to each other in one of 3 middle seats at tables. again, choose a table, choose ${3}\choose{2}$ seats, and order of Joe Frank, and then other remaining people: $|A_{mid}|$=3*2*${3}\choose{2}$*${13}\choose{8}$

Then answer is ${15}\choose{10}$ - $|A_{edge}|$-$|A_{mid}|$

c) similar to b, except now let A_across be that they sit across from one another. This can only happen if one sits at north table and the other at south. choose who sits at north table, then choose any of 5 seats (other must sit across from him), and then arrange remaining people: $|A_{across}|$=2*5*${13}\choose{8}$ Subtract this from answer to b)

I know the answers are wrong, I just don't understand why, nor what the answer should be.

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  • $\begingroup$ Welcome to MSE. Great question, but your MathJax could use a little work. Check this link. $\endgroup$ Jul 5 '18 at 9:13
  • $\begingroup$ Yeah you're right. I just edited it. Any other suggestions on how to make it neater? $\endgroup$
    – Adam G
    Jul 5 '18 at 9:16
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You basic error in all 3 answers is you are using combinations ${15} \choose {10}$ etc.

${15} \choose {10}$ would give you the number of ways of seating 10 people in 15 seats if the people were indistinguishable. But here the people are distinguishable - we know that Joe and Frank, for example, are identified by name. So an arrangement in which Frank sits at seat 1 and Joe sits at seat 2 is different from an arrangement where Joe sits as seat 1 and Frank sits at seat 2.

So in part (a), seating 10 distinguishable people in 15 seats, the first person has a choice of 15 seats, the second has a choice of 14 seats etc. and we have a total of $\frac{15!}{5!}$ arrangements.

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  • $\begingroup$ of course (face palm). So then for b) and c), would I just replace ${13}\choose{8}$ with $\frac{13!}{5!}$? $\endgroup$
    – Adam G
    Jul 5 '18 at 9:49

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