alphonsa wrote:
In how many ways can a group of 15 friends be seated round 2 tables if one of the tables can seat 10 and the other can seat 5 people?
A) 15C5 * 9! *4!
B) 15C3 * 8! *3!
C) 15C4 * 9! *3!
D) 15C2 * 10! *3!
E) 16C5 * 10! *4!
Source: 4gmat
Please explain..
This is a great question to test your understanding of counting methods. It tricks you into thinking that you'll have to do a lot of complex math, but if you apply a method I like to call "Half Baked" you can find the correct answer very shortly. (Half baked basically means only solve the problem to the extent you need to eliminate all of the wrong answers.)
First, I start by building the first table. I'm choosing 10 people from a group of 15: 15C10. However, none of my answer choices reflect this. So I apply another principle that lets me rephrase the answer. If you have
the Official Guide, look at the last sentence of the Counting Methods section (4.1.10) where it says nCk = nC(n-k). This makes sense because if I start with the OTHER table, I would have gotten 15C5 and ended up with the same scenario.
This means that 15C10 = 15C5.
Given that answer choice A is the only one that matches this start, I would select it and move on.
But you might be interested in knowing more about how to completely solve the problem.
Having established the first part of the answer, now I'm interested in knowing what order each person will sit in at the table.
So I have a table of 10 and a table of 5, and I need to order people around each table. The trap answer here is 10! and 5!. However, remember that in a circular table, there is a unique circumstance where it matters who is sitting on either side of the person, not necessarily what seat they occupy. Therefore, if everyone in the group shifts one seat to the left, then it is still the same table arrangement.
Instead of 10! and 5!, you account for the rotation around the table by reducing the factorial number by 1, getting 9! and 4! as the answer.
So your final answer becomes 15C5 * 9! * 4!.