A tea party is arranged for 16 people along the two sides of

I have a slightly different version :

8C4.10C4.8C2.6!

My soln (twixt , I hope I haven't messed this one up).

The 4 ppl on side choose 4 out of 8 seats , so do the other 2 ppl.
The 4 and 2 then permute amongst themselves.
The remaining occupy the remaining 10 seats and then permute among themselves.

So, my ans.
8C4.8C2.4!.2!.10!

HMTG

I got 8C4 * 8C2 * 10!
What is the OA?

my answer:8C4*4!*8C2*2!*10!

Guys I am totally away from your solutions

2*[7!*2!*5!*4!]

I get 2*(10 C 6)* 8! * 8!
What's the OA?

The problem involves two parts.

part 1 - Combinations:

4 want to sit on one paticular side - so this can only be done in one way.
2 want to sit on another side - this can be done in only one way.

That leaves 10. These 10 (first worry about breaking them into one side or the other. We will worry about seating - arrangements in Part 2 of the solution) can be divided into two groups (group of 4 and group of 6).

Out of 10, 4 need to go on one side (and the rest automatically go on the other side). So, this alone can be done in 10C4 (or if you work on the other side, you will end up with 10C6. Note that 10C4 and 10C6 are the same). So, part one of the answer is 10C4 (or 10C6 both are equal).

Now, the problem asks about howmany ways they can be seated.

Part 2 - Permutations:

As far as arranging them, the problem did not say the team of 4 has to sit together (the problem only states that the team of 4 want to sit on a particular side - together or not).

therefore, one side of the table can be seated in 8! ways and with the same logic the other side can be seated in 8! ways.

So, the answer should be 10C4 * 8! * 8!.

If the problem stated that the team of 4 should be seated on one particular side together and team of 2 shoudl be seated on another side together, then
the answer might be

10C4 * 5! * 7! * 4! * 2!.

if the problem stated that the team of 4 should be seated together on one side and team of 2 should be seated together on the other side (without saying a 'particular' side), then anwer is

10C4 * 5! * 7! * 4! * 2! * 2!.

-Srinivas

It is 10C4*8!*8!


Here is my solution

Let 4 persons sit on one side(I) and 2 on other side(II).

Remaining 10 are there to be seated out of which 4
have to be selected to be seated on (I) ==> 10C4 ways

rest of the six have to sit on other side in 6C6 ways=1

Again on either sides the can be arranging themselves in 8! ways each.

REMEMBER: Here the question never asked for the 4 persons and the 2 persons sitting together. Rather it said that they are to be on the same end.

So 10C4*8!*8!

See my earlier post.

Srinivas: Yes you were right....

Srinivas - I guess TYR reply is right. How come this a combination problem this is fully a permutation problem.

Rearrange 8C4 * 8C6 * 10! to see that it is indeed 8! * 8! * 10C6

-Srinivas

My apologies, read this as "rearrange (8!/4!)*(8!/6!)*10 to see that it is indeed 8! * 8! * 10C6"

([quote="Srinivas"]Rearrange 8C4 * 8C6 * 10! to see that it is indeed 8! * 8! * 10C6

-Srinivas

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.