1. ## Combination confusion

I am having real trouble figuring out when to use these two formulas:

c(k + t - 1, t-1) and c(k + t - 1, k)

where k is equal to number of selections
t is equal to number of elements; number of groups items are in I guess is another way to state that

For example:

There are piles of red, green and blue balls; each pile has at least 10 balls in it. How many ways can you select 10 balls?

I would think it would be c(10 + 3 - 1, 3-1) but the answer the book gives is c(10 + 3 - 1, 10)

Another one is:

How many ways can 10 be selected if exactly one red ball must be selected?
I get c(9 + 2 - 1, 2-1) but again the answer is c(10 +2 -1, 9)

You have a bag with twenty balls in it. six red six green and 8 purple

How many ways can we select 5 balls if balls of each specific color are identical?

i think this is 20!
______
6!6!8!

Could you tell me if this is correct? Thanks again folks very much.

2. $\binom {n}{k} = \frac{{n!}}{{\left( {k!} \right)\left( {n - k} \right)!}}=\binom {n}{n-k}$

3. ## Combination confusion

So what you are saying is that c(9 +2-1,9) is the same as c(9+2-1,2-1)????

4. Originally Posted by Frostking
So what you are saying is that c(9 +2-1,9) is the same as c(9+2-1,2-1)????
Well, of course.
$\binom {9+2-1}{2-1} = \frac{(9+2-1)!}{(2-1)!([9+2-1]-(2-1)])!}$

5. ## Combination confusion

PLato, I finally get it. It still does not seem like it should be true but I have tried several using either value k or the number you are choosing - 1 and they are equal. I do not know why it is so hard for me to grasp but I appreciate your patience. Maybe there is someone else out there that was also confused and this will help them before they get as frustrated as I did! Thank s again for the assistance.

6. Originally Posted by Frostking
I do not know why it is so hard for me to grasp . Maybe there is someone else out there that was also confused and this will help them before they get as frustrated as I did!
Did you ask for a combinational proof elsewhere?
Here is an example. If we have a group of ten from which we choose three then there are seven not chosen. On the other hand, if we choose seven then there are three not chosen. From n if we choose k then there are n-k not chosen and visa-versa.