Statistics permutations - how many ways to make n couples

[MaryStew]

Suppose we have a group of n women and n men, all heterosexual and all strictly monogamous. How many ways are there to make n couples out of these 2n people?

So apparently the given solution is n! which I don't quite understand.

In the textbook it says:

n=1 --> 1 couple

n=2 --> 2 couples

n=3 --> 6 couples

But I don't really get this logic. For example if n = 2, wouldn't the choices be (w1m1, w1m2, w2m1, w2m2) giving 4 choices instead of 2? I don't understand what 2 choices it's referring to. The only example I agree with is n = 1 where the only possible choice is w1m1.

 

[MarkFL]

Let's pull out 1 woman and let her choose a man. How many choices does she have? Next, we pull out another woman and let her choose from the remaining men...how many choices does she have? Continue in this manner, and use The Fundamental Counting Principle to determine the total number of possible couples.

 

[pka]

Suppose we have a group of n women and n men, all heterosexual and all strictly monogamous. How many ways are there to make n couples out of these 2n people?
So apparently the given solution is n! which I don't quite understand.
In the textbook it says:
n=1 --> 1 couple
n=2 --> 2 couples
n=3 --> 6 couples
But I don't really get this logic. For example if n = 2, wouldn't the choices be (w1m1, w1m2, w2m1, w2m2) giving 4 choices instead of 2? I don't understand what 2 choices it's referring to. The only example I agree with is n = 1 where the only possible choice is w1m1.

MaryStew, take note the in the book's example you have $n!,~1!=1,~2!=2~\&~3!=6$
Thus for five pairs we get $5!=120$.
Think of lining the $n$ men in strict alphabetical order.
Then in how many ways can the women form a line?
Is it $n!$? That gives us the answer.

 

[Deleted member 4993]

But I don't really get this logic. For example if n = 2, wouldn't the choices be (w1m1, w1m2, w2m1, w2m2) giving 4 choices instead of 2? I don't understand what 2 choices it's referring to. The only example I agree with is n = 1 where the only possible choice is w1m1.

When you are choosing w1m1, you have already fixed w2m2. In that case w2m2 is not a free choice anymore.

Then the other choice is w1m2 - again you have fixed w2m1

Thus you really have two choices (w1m1 and w1m2) other two gets "chosen" for you by constraint.

 

[Steven G]

Suppose we have a group of n women and n men, all heterosexual and all strictly monogamous. How many ways are there to make n couples out of these 2n people?

So apparently the given solution is n! which I don't quite understand.

In the textbook it says:

n=1 --> 1 couple

n=2 --> 2 couples

n=3 --> 6 couples

If n=1, then 2n = 2 and in the end there will be 1 couple (not the number of different ways to make the 1 couple which is 1!)
If n=2, then 2n = 4 and in the end there will be 2 couples (not the number of different ways to make the 2 couples which is 2!)
If n=3, then 2n = 6 and in the end there will be 3 couples (not the number of different ways to make the 3 couples which is 3!)
Do you understand now?