Math Is Fun Forum

jacks

Member

Registered: 2012-11-21

Posts: 132

no. of ways of dividing in a couple

the number of ways of dividing

men and

women into

couples each consisting of a man and a woman is

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: no. of ways of dividing in a couple

Hi;

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: no. of ways of dividing in a couple

hi jacks and bobbym,

Am I missing something here?  Why isn't it 15!

http://www.math.hawaii.edu/~hile/math100/combb.htm example 2.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: no. of ways of dividing in a couple

Hi Bob;

That does not pair the people up as (man, woman), this is not an arrrangenent but a selection problem.. Here is how I did it.

The first man can be partnered with 15 different women. The second man with 15 different women, all the way to the fifteenth man. 15 x 15 = 225.

But then again...

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: no. of ways of dividing in a couple

hi bobbym,

It looks like you've got the same woman dancing with two men?

This is what I did (before I researched it).

Get all 30 into the ballroom.  Expel everyone else. (except the band of course).

Line the men up against the wall.  (So they cannot escape if they don't like their partner )

Man one can have any of 15 partners.  Once chosen, man two can choose from 14, man three from 13 etc etc and man fifteen gets the last woman.

Let's simplify.

Say three men Al, Bob and Colin and three women, Di, Elaine and Fiona.

Al-Di; Bob-Elaine; Colin-Fiona
Al-Di; Bob-Fiona; Colin-Elaine
Al-Elaine; Bob-Di; Colin-Fiona
Al-Elaine; Bob-Fiona; Colin-Di
Al-Fiona; Bob-Di; Colin-Elaine
Al-Fiona; Bob-Elaine; Colin-Di

That's 3 x 2 x 1 = 6 possibilities.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: no. of ways of dividing in a couple

Hi;

I am forced to agree, I think I misread the question. My answer is for forming one dance couple, which would make this a committee problem. Hopefully, 15 x 15 is correct then. Now back to reality...

The actual problem:

the number of ways of dividing

men and

women into

couples each consisting of a man and a woman is

looks like it says 15 couples which makes your solution correct. Wunderbar!

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.