Number 27

[adam1111]

Does anyone know how to do that problem without making a systematic list???

[adam1111]

haha i figured it out

3 Choose 3 + 3 Choose 2 + 3 Choose 1 = 10
It came to me in a dream :)

Now how exactly do you do bearings i # at drawing diagrams

[collegebookworm]

Bearings are easy. Think of them like hands on a clock: they start at 12pm and move clockwise. Therefore, 199° would be pointing in the general southwest direction.

If it's the calculating bearings from other bearings you don't get, you need to better yourself at drawing diagrams and finding component vectors.

[collegebookworm]

Well, that's because we need to do 5 C 3 = 10, because we have 3 options, and three spots to fill.

Or, conversely, 5 C 2 works just fine as well

[collegebookworm]

No, you misunderstand me. Say you had 5 options and wanted to fill 8 spots. Then you would have 12 C 7 or 12 C 5

[collegebookworm]

*cough, cough* check your answer *cough cough*

[stupidityismygam]

[bradp]

worlow, I remember in the calc room @ state last year you saw that I had the balls and bins thing written down on a piece of paper.

You were like "Duuuuuuuuuuuude, you're so lame, balls and bins doesn't show up on uil tests"
and i was like "dude, don't be so mean"

[stupidityismygam]

[mathslug]

Wait a minute here...

Adam, I'm not sure if doing 1 to N combinations on the number works:

3C3 + 3C2 + 3C1 = 1 + 3 + 3 = 7

I was wondering if anyone has generalized a formula yet... I've been toying with this problem inductively for a while, and I have a method for solving it, but it feels wrong. The way I do it is the following...

Add one to start with. This is our base case where everything is different to start with. "One of each" if the number of things you're working with here is not one or less... then proceed to the next step.

Add N, where N is the number of things you are working with. This will take care of each case where you only get one kind of thing in your set. If the number of things you are working with is not 2, then proceed to the next step.

add N(N+1)/2!, i.e. sum of numbers 1 to N. This takes care of our first round of duplicate cases. If the number of things you're working with is more than 3 then continue to the next step.

add N(N+1)(N+2)/3! , and if the number of things you're working with is more than 4 then continue in likewise fashion.

I'm not sure if this method works correctly or not. I still havent proven it. I got it from careful inspection of the characteristics of each generarted set given 1-4 elements. I know it works for 1-4, but I'm hoping that someone can prove it for more than that.

For 3:

1 + N + N(N+1)/2 = 1 + 3 + 3*4/2 = 10

For 4:

1 + N + N(N+1)/2 + N(N+1)(N+2)/6 = 1 + 4 + 4*5/2 + 4*5*6/6 = 35

For 5: ?

1 + N + N(N+1)/2 + N(N+1)(N+2)/6 + N(N+1)(N+2)(N+3)/24 = 1 + 5 + 5*6/2 + 5*6*7/6 + 5*6*7*8/24 = 1+5+15+35+70 = 126?


Can anyone help? And if my method is wrong, how should I go about solving this in the future?

Zack

[mathslug]

Oh, I apologize. I did not see the Balls in Bins (kinda awkward to write) post. Argh... That's WAY easier than what I was doing. Can you explain why this works?

Zack

[stupidityismygam]

[Celsion]

[stupidityismygam]