Question

twelve people arrive at restaurant. there is one table for six, one table for 4 and one table for 2.

In how many ways can they be assigned to a table

Answer 1

good morning imran
just did this problem, but i forget about the circle

Answer 2

this isn't a circle problem though

Answer 3

Since total Head count is 12. So for 1st table for six the probability is 6*12=72
Now only six is left and so for 2nd table for four the probability is 4*6=24
Now only two is left and so for 3rd table for 2 the probability is 2
so total no. of ways is 72+24+2=94

Is it correct?

Answer 4

no that is not correct

Answer 5

no the answer is 13 860

Answer 6

http://openstudy.com/groups/mathematics/updates/4e5f86dd0b8b1f45b49e5b11

Answer 7

what i did was :

12 C 6 + 12 C 4 + 12 C 2

Answer 8

but its wrong lol

Answer 9

oh wait i should learn to read. it doesn't say how many arrrangements are there, just how many possible assignments

Answer 10

12 people 12 chairs

Answer 11

the circle arrangement doesn't apply to this question imran

Answer 12

should be
\[\dbinom{12}{6}\times \dbinom{6}{4}\]

Answer 13

cause it doesn't state that the tables are circuar

Answer 14

no i said that and i was wrong

Answer 15

can you explain it to me sattellite

Answer 16

first off is the answer 13860?

Answer 17

yes it is

Answer 18

It does not say the tables are separated or put together just says a table that holds 6, 4 and 2 which would get your 12 chairs and you have 12 people now how many combinations can you create

Answer 19

because i don't want to start explaining something that is wrong!

Answer 20

lol no ur correct

Answer 21

ok so we first have to be clear about what the problem is asking. it wants the number of ways you can partition 12 people in to one group of 6, one group of 4 and one group of 2

Answer 22

yep

Answer 23

Ok so the tables are separated

Answer 24

pretty sure they are

Answer 25

out of 12 people you are going to choose 6 to be at table A. how many ways are there to do this?
the answer is
\[\dbinom{12}{6}\] which really just restates the question. we compute
\[\dbinom{12}{6}=\frac{12\times11\times 10\times 9\times 8\times7}{6\times 5\times 4\times 3\times2}=924\]

Answer 26

that is the number of ways to choose 6 from a set of 12

Answer 27

now you have 6 people left and you want the number of ways to assign them to table B
that is
\[\dbinom{6}{4}=\frac{6\times 5}{2}=15\]

Answer 28

notice you could have said "how many ways to assign them to table C?" and got
\[\dbinom{6}{2}\] which is also 15

Answer 29

now you are done because once you have assigned 6 to table A and 4 to table B you have no choices for table C

Answer 30

so 924 ways to make up table A, 15 ways to make up table B, then only one way for table C and by counting principle the number of ways to do this all together is
\[924\times15 =13860\]

Answer 31

thanks so much for the explanation:)

Answer 32

yw hope it is relatively clear