1. ## 2 Counting problems

1.In how many ways can 12 persons seat themselves at 3 different round tables, 4 at each:

(a) (12!) / (4!)^3
(b) (12!)(3!^3) / (4!^3)
(c) 12!
(d) none of the above

2. In how many ways can 12 persons be divided into 3 groups of 4 each:

(a) (12!) / (4!)^3
(b) (12!)(3!^3) / (4!^3)
(c) 12!
(d) none of the above

2. ## Re: 2 Counting problems

Originally Posted by swordfish774
1.In how many ways can 12 persons seat themselves at 3 different round tables, 4 at each:
(a) (12!) / (4!)^3 (b) (12!)(3!^3) / (4!^3) (c) 12!
(d) none of the above

2. In how many ways can 12 persons be divided into 3 groups of 4 each:
(a) (12!) / (4!)^3 (b) (12!)(3!^3) / (4!^3) (c) 12!
(d) none of the above
These are very nice problems.
Do you plan to show any effort towards solving them?
You know that this is not a homework service?

3. ## Re: 2 Counting problems

you are the second user to say that. (I posted another question, someone else said the same thing) I am here to just verify the answers and look for better methods

1.
selection for tables
12C4*8C4*4C4 = (12!) / (4!)^3

then you muliply 3! for circular permutations. so option b

2. 12C4*8C4*4C4 = (12!) / (4!)^3

then divide by 3! so option 4

I came here to verify answers. and 2. question I am not sure because last option is 'none of these', so need to check if approach left some counts or if there was some double counting.

4. ## Re: 2 Counting problems

oo. Plato... you are the same user.