Question

Permutation and Combination
In a parent-teacher association, there are 5 couples. A project team of 4 people us going to be selected from the 5 couples. Find the number of ways to form the team if:

(a). the team consists of no couples;
(b). the team consists of exactly 1 couple.

Thanks, please explain briefly!

Answer 1

anyone know?

Answer 2

Yikes, this involves words as well as equations... I hope you're ready for this XD

Answer 3

yep

Answer 4

Well I'm not. Give me a moment to recall how to do this, shouldn't take more than three minutes...

Stand by....

Answer 5

haha.. okay

Answer 6

Okay, I think I'm ready. Commencing:

Answer 7

Now, this is all about the principle of multiplication, isn't it?
It's all about choices, and the number of choices for each position.
Allow me to illustrate what I mean.

Suppose we are allowed to make a team of four people, with no restrictions whatsoever.
I've drawn up this convenient little box to make it simpler to understand the concept:
\[\Large \left[\begin{matrix}?&?&?&?\\\text{I}&\text{II}&\text{III}&\text{IV}\end{matrix}\right]\]

Now hang on, this isn't yet the gist of it. In a team of four, you can arrange them in 24 different ways which doesn't really make a difference to the team, IE,a team consisting of ABCD is the same as the team consisting of CDBA. To avoid this overcounting error, we must also divide the product by 24.

Ready?

Answer 8

The answer of (a) is 80, and the answer of (b) is 120.

Answer 9

I think the step of (a) should be \[5C1 * 4C1 * 2 *2\] , but why should I *2 twice?

Answer 10

There are many different ways to approach counting problems such as these, and they should all lead to the same correct answer if all methods are logical. Unfortunately, I do not know of this method you have just posted. That is probably why you are tasked to explain the method even briefly, just so the checker is certain that you are not randomly spouting ways to acquire the correct answer (80). That being said, it doesn't look like you know the logic behind this solution you posted, and sadly, neither do I :(

Answer 11

Okay, and speaking of methods, I have just found a more elegant solution... one you may well be more comfortable with.... ready?

Answer 12

@y1066380
You have to work with me here :)

Answer 13

Ok

Answer 14

Okay, so, we want no couples at all in this four-player team.

So, that means the four-player team will have members of four different couples, yes?

Answer 15

Yes

Answer 16

Effectively, and conceptually, this is so far equal to \[\Large _5C_4\] right?

Answer 17

Yep, choose 4 couples from 5.

Answer 18

Okay, so now that we have four couples, for each of those couples, we have two choices right?
So includes a factor of \[\large 2\cdot 2\cdot 2\cdot 2=2^4\] right?

Answer 19

for each of those couples, we have two choices << you mean either M or F?

Answer 20

I was avoiding M or F in attempting to be sensitive of the possibility of a gay couple, but okay XD

yes, M or F :)

So two possibilities each, yes?

Answer 21

haha yes, i think i know it

Answer 22

5C4 * 2^4 ?

Answer 23

Precisely.
Compute.

Answer 24

Yep, then how about (b)?

Answer 25

You get 80, right?
Good.
Now b.

A four-player team with *exactly* one couple. It'd be prudent to FIRST pick THE couple that would be in the team, aye?

So, this can be done, with five couples in exactly \[\Large _5C_1\] ways, agreed?

Answer 26

yes

Answer 27

Now, from the remaining four couples, we need to pick two members who are not of the same couple. Easy enough. From the idea in the previous item, simply pick the two couples from which these two members originate. This can be done in \[\Large _4C_2\] ways, correct?

Answer 28

yep

Answer 29

Oh,,, 5C1 * 4C2 * 2 *2 , right?

Answer 30

Very good. 2*2 comes from the M or F choices for the latter two members (the ones who mustn't be of the same couple)

Solve, and you should get your desired answer ^_^

Answer 31

Great! thank you so much!!!

Answer 32

No problem. Combinatorics are something of a hobby of mine ^_^