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# Counting Chairs

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There are 8 dentical chairs around a circular table. In how many ways can 5  people be seated in 5 of these chairs if Richard wants to sit next to either Surya or Tamas (or both Surya and Tamas)?

(Consider two seating arrangements to be the same if one seating arrangement can be rotated to obtain the other.

So I thought about assigning the people letters so I could refer to them easier.

RST
RSA
RTA
RSB
RST

There are some possible combinations^
But i am having trouble finding the rest

Aug 8, 2022

#1
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There are $$8! \div 8 = 5040$$ ways to seat the 8 people (divide by 8 to account for rotations).

Let's use complementary counting for this problem, as it avoids counting multiple cases. So, we are looking to find all the failures (situations where Richard isn't sitting next to either person)

There are 8 spots for Richard to sit in, 5 spots for the person to the left of Richard (can't be S, T, or R), and 4 spots for the person to the right of Richard (can't be S, T, R, or the person to the left of Richard). Lastly, there are $$5! = 120$$ ways to seat the other 5. However, we have to divide by 8 (to account for rotations), so there are $$8 \times 5 \times 4 \times 5! \div 8 = 2400$$ ways to "fail".

This means that there are $$5040 - 2400 = \color{brown}\boxed{2640}$$ successes.

Aug 8, 2022
#2
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Why cannot it not be S or T?

Aug 8, 2022
#3
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We are using complementary counting, which means we count for everything we don't want.

In this case, we are counting all the ways we can have Richard not sit next to S or T, so we restrict the number of people who can sit next to Richard.

Hope this helps!

Aug 8, 2022
#4
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But we want richard to sit next to surya and or tamas

Aug 8, 2022
#5
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Yes, at the end of the day, that is what we are counting for, but we count all the ways to "fail" and subtract that from the total ways.

For example, here's a simpler problem: You flip a fair coin twice. How many ways can you get at least 1 head?

For this problem, instead of counting the ways we can get a head, we count all the ways we don't get a head. This can happen in 1 way (back-to-back tails), and there are 2 x 2 = 4 total ways to flip the coin, so there are $$4 - 1 = 3$$ ways to get at least 1 head.

We are doing the same thing here, just that it is a bit more complicated. If we subtract all the failures(where Richard does not sit next to S or T) from the total ways to seat the 8 people, we are left with the number of ways where Richard sits next to S or T.

Aug 8, 2022
#6
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Hmm. I understand. I tried 2640 but that is incorrect?

Aug 8, 2022
#7
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I don't see where you could have went wrong but 2640 is incorrect

Aug 8, 2022
#8
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Hmm... I don't know either...

BuilderBoi  Aug 8, 2022