Some Math

[mister_s]

Let me ask a simple math question that I'm having a disagreement about:

If I have 6 different CDs, how many different ways can I distribute the CDs so that 3 people receives exactly 2?

[VerticalDiscussions]

Just 1!! Its not like you can give 1 + 0.5 CD's to someone. Stupid maths...

 

 

?

[MoonlightSylv]

One.... duh

[princedylan101]

15*14*13 = 2730

[Canada EH]

OK so you got 1-6 cd's and a-c people.

54 different ways.

 

A gets 1,2 - 5,6 - 3,4

B gets 3,4 - 1,2 - 5,6

and C gets 5,6 - 3,4 - 1,2

 

next its A gets 1,3 so on so forth

 

so times 9 for the first round by 6, so 9x6=54.

Might be a differenet equation for it.

Cant see that happening the numbers are too high, I can envision 54 possibilities for sure.

[Bsmith]

4 hours ago, Canada EH said:

OK so you got 1-6 cd's and a-c people.

54 different ways.

 

A gets 1,2 - 5,6 - 3,4

B gets 3,4 - 1,2 - 5,6

and C gets 5,6 - 3,4 - 1,2

 

next its A gets 1,3 so on so forth

 

so times 9 for the first round by 6, so 9x6=54.

Might be a differenet equation for it.

Cant see that happening the numbers are too high, I can envision 54 possibilities for sure.

 

that's one way to do it, but dependant on the subject being covered the answer might still be wrong.

[Canada EH]

Its just variations to each having 2 cd's.

I would have no clue what the equation would be.

The example i did was one variation, equaling 9 different ways the cd's could be spread out.

I think you would have to figure out all the ways one person could recieve all 6 cd's, then expand it.

For example person A, can get 1,2 as in my example, but then 1,3 - 1,4 - 1,5 - 1,6 then 2,1 would now be valid, but 2,3 - 2,4 - 2,6

then 3,1 and 3,2 would not be valid, but these would be 3,4 - 3,5 - 3,6

4,1 - 4,2 - 4,3 are all not valid but these would be 4,5 - 4,5

5,1 - 5,3 - 5,4 not valid but these would be valid 5,2 - 5,6

and finally 6,1 - 6,2 - 6,3 - 6,5 and this would be 6,4

 

that is one variation - do that for each of the remaining 2 people.

 

But yeah I hear ya, its all about the wording because there would always be over-lap. A more interesting wording would be if people can only get 2 cd's but what are the variations if you put them in a hand. Cant not have the same cd in the same hand.

 

I will think about it more and expand the numbers to the remaining 2 people.

BTW O.P. What grade and area of math is this?

I kinda of like math questions, but most of the time find this line of forum post a little odd. For us to help you with your homework.

This question interests me. Would love to know what the equation is though.

As of now though I gotta get kicking and do some shit today.