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one of my friend asked me this question
there are infinite numbers of ball of three kind i.e. ball A,B,C. I mean there are infinite number of ball of kind A and infinite number of ball of kind B an so on.the problem is that one have to select 10 balls out of these balls,in how many "ways" he can do this? my approach for this problem was:

i can select first ball in 3 ways (A,B,C) and second ball in three ways and so on.I have 3 choices for every ball i am 3*3*3*3*3*3*3*3*3*3 total number of ways of doing it. but my friend argues that selecting (A,B,A,B,A,B,A,B,A,B) AND (B,A,B,A,B,A,B,A,B,A) should be counted once and similar for other sets of balls. but as question says i have to find number of "WAYS" of doing it,so both the cases are different. as i should focus on "how" to select not "what" to select to find the number of "ways". please correct me if i am wrong.thanks a lot for paying attention on this problem of mine.

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marked as duplicate by Ross Millikan, Amzoti, Alex Wertheim, Chris Godsil, Micah Jun 5 '13 at 1:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

If order matters, then the answer is $3^{10}$ as you observed. If order doesn't matter, then the problem amounts to finding the number of elements in the set $\{(a,b,c) \in \mathbb{N}^3 \,|\,a+b+c=10\}.$ I don't know the answer in this case. Hope that helps. Edit. I think the answer to this latter problem is $55$. – goblin Jun 4 '13 at 20:11
@user18921 thanks for paying attention,i understand how you are suggesting 55.but it is the language which is bothering me.i mean order matters or not in above question? – Abhishek Pandey Jun 4 '13 at 20:16
Sorry I don't understand. If order matters, that's one problem. The answer is $3^{10}$. If order doesn't matter, that's another problem, I think the answer is $55$. Since the word 'ways' does not have a technical meaning, that's the best we can do. – goblin Jun 4 '13 at 20:18
@user18921 i got it thank you – Abhishek Pandey Jun 4 '13 at 20:21
So, i'm interpreting that the question is asking 'How many solutions are there to: a + b + c = 10'. Maybe we can get somewhere with representing this with binary numbers (read stars and bars) where the # of 0's before the first 1 is the amount of a's, the # of 0's before the second 1 is the amount of b's, and the # of 0's after the second 1 is the amount of c's. For example, 000000000011 means there are ten a's, 0 b's, 0 c's. 000100000001 means there are 3 a's, 7 b's, and 0 c's. There are C([10+2], 2) ways to arrange those 1's. – Ozera Jun 4 '13 at 20:22
up vote 2 down vote accepted

I'm just repasting my answer from the comments section.

Interpreted as: "How many solutions are there for: a+b+c = 10".

Using the concept of 'star and bars' we write a few examples in binary numbers in which the # of 0's before the first 1 is the # of a's, # of 0's before the second 1 is the # of b's, and the # of 0's after the second 1 is the number of c's. Note, in general there are n-1 1's (bars) and k 0's (stars).


000000000011 corresponds to 10 a's, 0 b's, and 0 c's.

000001000001 corresponds to 5 a's, 5 b's, and 0 c's.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

The goal is to now figure how many ways we can arrange our 1's (or similarly arrange our 0's).

Note: Order does not matter. 5+5+0 = 10 is the same thing as 0+5+5 = 10

We can arrange our dividers in: $12 \choose 2$ ways


We can arrange our stars in: $10+3-1 \choose 10$


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