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42 votes
A teacher takes her AP calculus class of 8 students to lunch. They sit around a circular dining table.

a. How many seating arrangements are possible?
b. How many seating arrangements are there if the teacher has to sit on the chair closest to the soda fountain?
c. Among the students are one set of triplets. How many seating arrangements are there without all three of them sitting together?

User Ben Cochrane
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1 Answer

28 votes
28 votes

Explanation:

it is 8+1 people (8 students and 1 teacher).

a.

how many seating arrangements are possible ?

start with any of the 9 seats. there are 9 possibilities who will sit in that chair.

the next seat has then only 8 possibilities left. the next one 7. the next b one 6. and so on.

so, all possibilities are

9×8×7×6×...×1 = 9! = 362,880

b.

if the teacher is fixed to a specific seat, then this position is constant in all seating arrangements.

that means for the variable part we have only 8 people remaining.

and that gives us now only

8! = 40,320

possibilities

c.

how many combinations are there for the triplets, when they sit together, to arrange their internal sitting sequence ?

3! = 6

and we need to block these 6 combinations for each of the freely selectable seats as starting seat for such a block of 3.

to visualize this, let's give each seat an identifier.

we have the seats s1, s2, s3, ... s8, s9.

for all the combinations of these 9 seats, we need to eliminate the 6 combinations, where the triplets sit on s1, s2, s3.

and then the 6 combinations, where they sit at s2, s3, s4.

and then at s3, s4, s5. then at s4, s5, s6. ... s8, s9, s1. and s9, s1, s2.

it is not clear if b. still applies here (and the teacher has a fixed seat), or if we are back to the general a. scenario.

if we have again all 9 people picking their seats freely, and we only block the triplets from sitting together, then we need to block these 6 combinations for each of the 9 seats (as visualized above) giving us

9! / (9×3!) = 362,880 / 54 = 6,720 remaining possibilities.

if we have again the teacher blocking one fixed seat, then we need to block the 6 triplets combinations theoretically for each of the 8 remaining seats as above.

but : the teacher is now a separator disabling some of these blocks of 3.

e.g. let's say the teacher's seat is s9. then the teacher automatically separates the triplets in otherwise "illegal" blocks of 3.

s7, s8, s1 is now a valid seat combination for the triplets (because there is the fixed s9 in between).

the same for s8, s1, s2.

so, we need to eliminate the 6 combinations not for all 8 starting seats, but only for 6.

and that gives us

8! / (6×3!) = 40,320 / 36 = 1,120 remaining possibilities.

User Jason Marshall
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