In how many ways can a group of 15 friends be seated round 2 tables if

Question

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..

LighthousePrep · Accepted Answer

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!.

Ashishmathew01081987 · Answer

IMO ans is A

Explanation

10 people can be selected from 15 people in 15C10 ways.
Remaining 5 people can be selected from 5 people in 5C5 ways.

Now, arranging 10 people on a round table = (10-1) ! = 9!
Arranging 5 people on a round table = (5-1) ! = 4!

Hence, total number of ways = 15C10 * 9! * 5C5 * 4! 
= 15C5 * 9! * 4!

15C10

Ashishmathew01081987 · Answer

In general the simple formula for arranging 'n' things or persons around a circular table can be summarized as follows

1) Number of ways of arranging 'n' people on a circular table = (n-1) !
2) When clockwise or anticlockwise observation are not different then number of circular arrangements of 'n' different people = (n-1)! / 2

3) Number of selection of 'k' consecutive things out of 'n' things in a circle 
= n when k<n
=1 when k = n

Vivekksuri · Answer

Couldn’t understand why did we do 15C10*5C5

Posted from my mobile device

bumpbot · Answer

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In how many ways can a group of 15 friends be seated round 2 tables if

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