Thread: 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

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?

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.

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