# Combination and Selection Questions

• Sep 12th 2010, 05:20 AM
Lukybear
Combination and Selection Questions
Six married couples are in a room
How many ways can 4 people be randomly chosen if:
a) The four people chosen represent two married couples
b) Exactly one married couple is among the four people chosen

a) 6C2 = 15 ways.
b) Here is where I became stuck: total choice of 4 people are:
6C1 x 10C2

However, this include arrangements of couples
To get rid of double couples, 6C1 x 10C2 - 6C2 to obtain 255 ways

The book however has 270, by 6C1 x 10C2 - 6C1 x 5C1

Also, when doing b) why is it wrong to do this way:
First select the couple to be in the group. Then multiply this by no. of people available. Then multiply by no. of people left, which does not belong to couple pair.

6C1 x 10 x 8
• Sep 12th 2010, 06:23 AM
undefined
Quote:

Originally Posted by Lukybear
Six married couples are in a room
How many ways can 4 people be randomly chosen if:
a) The four people chosen represent two married couples
b) Exactly one married couple is among the four people chosen

a) 6C2 = 15 ways.
b) Here is where I became stuck: total choice of 4 people are:
6C1 x 10C2

However, this include arrangements of couples
To get rid of double couples, 6C1 x 10C2 - 6C2 to obtain 255 ways

The book however has 270, by 6C1 x 10C2 - 6C1 x 5C1

(a) agree
(b) Shouldn't that be 240 instead of 270? I get the same expression:

The number of ways to choose the couple is 6C1. Multiply this by the ways to choose the other two people, which is 10C2 - 5C1. It's 10C2 without restriction, and then you subtract off those that form a couple since this is disallowed. The book answer just expanded 6C1(10C2 - 5C1) by distributive property. (Or maybe they reasoned slightly differently but to the same effect.)

Quote:

Originally Posted by Lukybear
Also, when doing b) why is it wrong to do this way:
First select the couple to be in the group. Then multiply this by no. of people available. Then multiply by no. of people left, which does not belong to couple pair.

6C1 x 10 x 8

This is wrong because for the last two people you are making order matter, thus you are counting duplicates. In fact you count every second pair of people twice, thus your final answer is twice that of the correct answer.
• Sep 12th 2010, 06:35 AM
Plato
Quote:

Originally Posted by Lukybear
Also, when doing b) why is it wrong to do this way:
First select the couple to be in the group. Then multiply this by no. of people available. Then multiply by no. of people left, which does not belong to couple pair. 6C1 x 10 x 8

Quote:

Originally Posted by undefined
This is wrong because for the last two people you are making order matter, thus you are counting duplicates. In fact you count every second pair of people twice, thus your final answer is twice that of the correct answer.

But notice that $\dfrac{6\cdot 10\cdot 8}{2}=240$.
In other words, in this way of selecting the one couple & one non-couple each non-couple is counted twice.
• Sep 12th 2010, 06:57 AM
undefined
Quote:

Originally Posted by Plato
But notice that $\dfrac{6\cdot 10\cdot 8}{2}=240$.
In other words, in this way of selecting the one couple & one non-couple each non-couple is counted twice.

I thought that's what I said. :p
• Sep 13th 2010, 05:01 AM
Lukybear
Ok thanks. Can I just ask, the reasoning for 10C2 - 5C1 is that in the no. of ways of selection 2 people from 10, is included the no. of ways of selecting 1 couple from 5?

Also, when do you know you are making order matter and when no?

For example in this question (sorry if I am doubling question in one threat, but it is most appropriate this way)

12 People at restruant. One table for 6, one table for 4 and one table for 2. How many ways can table be assigned.

Solution is 12C6 x 6C4 x 2C2

Why isnt it divide by 3!?
• Sep 13th 2010, 07:35 AM
undefined
Quote:

Originally Posted by Lukybear
Ok thanks. Can I just ask, the reasoning for 10C2 - 5C1 is that in the no. of ways of selection 2 people from 10, is included the no. of ways of selecting 1 couple from 5?

Yes. You can think of it enumeratively and it might make more intuitive sense; 10C2 corresponds to a list of all the pairs of people out of those 10 people; 5C1 corresponds to a list of all couples out of those 10 people. The list we want to count is the first minus the second.

Quote:

Originally Posted by Lukybear
Also, when do you know you are making order matter and when no?

If you say, number of ways to choose the first and then number of ways to choose the second, you're making order matter.

Quote:

Originally Posted by Lukybear
For example in this question (sorry if I am doubling question in one threat, but it is most appropriate this way)

12 People at restruant. One table for 6, one table for 4 and one table for 2. How many ways can table be assigned.

Solution is 12C6 x 6C4 x 2C2

Why isnt it divide by 3!?

Because those tables are distinct.

Suppose we have 6 people to be assigned to 3 tables, each table seats 2. If the tables are labeled (distinct) then the number of ways is 6C2 * 4C2 * 2C2. Otherwise it's (6C2 * 4C2 * 2C2) / 3!.