Hi everyone. I'm currently studying basic combinatorics in college and I came across a question I'm having some trouble with:

Part 1: k number of b***s are thrown into 10 boxes - seven green, three red. More than one ball can be placed in each box. The way the course writes this is D(10,k). I'm asked to write a summation identity for this problem in the following format:

Part 2: Test the identity for k = 3.

Can anyone help me out with this? I've tried writing several identities and none of them work for k = 3. I believe the number of combinations I'm supposed to get is 220 and have been unsuccessful in writing an identity that uses summation to calculate that number.

Thanks!

narchitect May 25, 2015

#2**+5 **

Hi Narchitect, some more info regarding this problem would be useful.

Is there a maximum value for k ?

What do the 7 green boxes and 3 red boxes have to do with it ? That doesn't seem to affect the problem at all, that is, 10 boxes all of the same colour would appear to produce the same problem.

Getting the answer 220 for k=3 is easy enough and that seems to give some insight into how to proceed. However, it opens up questions such as, suppose, for example, k=9, then we could put 2 b***s in one box and the other 7 in 7 separate boxes, or we could have two boxes each containing 2 b***s with the other 5 in separate boxes, or three boxes each containing two b***s or 4 boxes each containing 2 b***s or one or two or three boxes each containing 3 b***s or ..... , 1 or 2 boxes each containing 4 b***s, and so on. Is there a function, or does there exist a function that will cover all of those possibilities ? What sort of function have you been trying ? Would a piece of computer code (an algoritm) be allowable as a function ?

Guest May 27, 2015

#1**+5 **

I don't know how to do this problems BUT are the b***s identical or different from one another?

Melody May 26, 2015

#2**+5 **

Best Answer

Hi Narchitect, some more info regarding this problem would be useful.

Is there a maximum value for k ?

What do the 7 green boxes and 3 red boxes have to do with it ? That doesn't seem to affect the problem at all, that is, 10 boxes all of the same colour would appear to produce the same problem.

Getting the answer 220 for k=3 is easy enough and that seems to give some insight into how to proceed. However, it opens up questions such as, suppose, for example, k=9, then we could put 2 b***s in one box and the other 7 in 7 separate boxes, or we could have two boxes each containing 2 b***s with the other 5 in separate boxes, or three boxes each containing two b***s or 4 boxes each containing 2 b***s or one or two or three boxes each containing 3 b***s or ..... , 1 or 2 boxes each containing 4 b***s, and so on. Is there a function, or does there exist a function that will cover all of those possibilities ? What sort of function have you been trying ? Would a piece of computer code (an algoritm) be allowable as a function ?

Guest May 27, 2015