# Thread: Question on Combinatorics

1. ## Question on Combinatorics

I'm learning for my highschool finals (in the Netherlands) and reviewing all the chapter I've done this and last year in maths.
For three weeks I'll be repeating Combinatorics and Probability Theory. And I have a question my math teacher couldn't answer.

Let's assume we need to make a row of 5 pencils. And those pencils can be black or white.
The ammount of colour combinations I can make is 2^5

Now suppose we can choose out of three kinds of colours: white, black and green.
The ammount of colour combinations is 3^5

Now let's say there must be three black pencils and two white ones
The ammount of colour combinations is:

5 ncr 3 or 5 ncr 2
( with ncr this is what I mean: 5 ncr 3 - Wolfram|Alpha)

Now my question:
Let's assume we have 3 black penciles, 5 white peciles and 2 green one. How many combinations can I make?

I know you can write all possibilities down but I'm curious if there is a mathematical way and if not why not.

2. ## Re: Question on Combinatorics

Originally Posted by MuddyMudskipper
Now my question:
Let's assume we have 3 black penciles, 5 white peciles and 2 green one. How many combinations can I make?
I know you can write all possibilities down but I'm curious if there is a mathematical way and if not why not.
Think of alpha-string $\displaystyle BBBWWWWWGG$ there are $\displaystyle \frac{10!}{3!\cdot5!\cdot 2!}$ ways to rearrange that string.

3. ## Re: Question on Combinatorics

@Plato - I believe the question is concerned with only strings of length 5. Here is my (unsatisfactory) solution to the problem.

We do not need to write out each possible unique 5 character string, we only need 1 example of each and then we can use counting methods. So to get an example of each, consider the fact that for each string we can either have 0,1,2, or 3 b's, then given how many b's we can either have 0,1, or 2 greens.

So here are the examples of each string:
WWWWW
WWWWG
WWWGG
BWWWW
BWWWG
BWWGG
BBWWW
BBWWG
BBWGG
BBBWW
BBBWG
BBBGG

Now we count the unique permutations of each of these and sum to get $\displaystyle \frac{5!}{5!}+\frac{5!}{4!}+\frac{5!}{3!\cdot 2!}+\frac{5!}{4!}+\frac{5!}{3!}+\frac{5!}{2!\cdot 2!}+\frac{5!}{2!\cdot 3!}+\frac{5!}{2!\cdot 2!}+\frac{5!}{2!\cdot 2!}+\frac{5!}{3!\cdot 2!}+\frac{5!}{3!}+\frac{5!}{3!\cdot 2!}$

Like I said, I am definitely not happy and would love to see a more elegant solution, but this sum is equal to 181

4. ## Re: Question on Combinatorics

I know I at least think that I've "solved" this question, but can anyone see a more elegant way of counting this up? I feel like mine is stupid.

5. ## Re: Question on Combinatorics

Originally Posted by MonroeYoder
I know I at least think that I've "solved" this question, but can anyone see a more elegant way of counting this up? I feel like mine is stupid.
I am not sure that there is a simpler way than yours to solve that question if it indeed is way you read it.
I am not convinced that is a correct reading.
Here is my reason for thinking that: "that only complicates the solution"
It teaches nothing new. Whereas, my reading brings a fundamentalism principle into play.
Now yours may well be the correct reading of the question.
But if it if then in my view it nothing more that a busy work problem.

6. ## Re: Question on Combinatorics

I completely agree that this type of problem is "busy work." As someone who has taken Combinatorics in college I am somewhat distressed that I can't come up with a better way. I mean if we have
x B's, y W's and z G's,
and we want to know how many unique strings of length $\displaystyle n$ we can make with the characters B W and G, it seems like there should be a general technique to do so.

Thanks for giving it consideration though.

7. ## Re: Question on Combinatorics

I am not an expert here, but why is this not a simple problem of combinations? We have 10 items taken 5 at a time
C(10,5) = 252?