# Thread: How many unique combinations in this scenario?

1. ## How many unique combinations in this scenario?

Hello everyone. I need some help answering this question. I have 16 colors to choose from, and I can arrange any 15 in a row at my choosing, but the following example would not count as two unique combinations, since the end result is the same color (14 reds and 1 blue).

Locations 1-14 are Red and location 15 is Blue
Location 1 is Blue and locations 2-15 are Red

A factorial would count these as two combinations, so I am looking for the right equation to solve this. How many UNIQUE color combinations are possible?

Thanks!

2. ## Re: How many unique combinations in this scenario?

Do you need the number of multisets with 15 elements taken from the set of 16 elements? That would be $\displaystyle \left(\!\!{16\choose 15}\!\!\right)$ (see the section "Counting multisets" in the link above).

3. ## Re: How many unique combinations in this scenario?

Sort of. I need the number of unique color combinations I can create by selecting 15 colors from a set of 16 available colors. I think your above equation does not account for my example: These two would not count as two unique combinations, since the end result is the same blended color (14 reds and 1 blue).

Locations 1-14 are Red and location 15 is Blue
Location 1 is Blue and locations 2-15 are Red

Make sense?

4. ## Re: How many unique combinations in this scenario?

Originally Posted by MathinTExas
I have 16 colors to choose from, and I can arrange any 15 in a row at my choosing, but the following example would not count as two unique combinations, since the end result is the same color (14 reds and 1 blue).
Locations 1-14 are Red and location 15 is Blue
Location 1 is Blue and locations 2-15 are Red.
I find it hard to follow that example.
Is the arrangement of beads 1-9 & 11-15 being red and bead 10 blue the same as the above?.

5. ## Re: How many unique combinations in this scenario?

Originally Posted by MathinTExas
I think your above equation does not account for my example: These two would not count as two unique combinations, since the end result is the same blended color (14 reds and 1 blue).

Locations 1-14 are Red and location 15 is Blue
Location 1 is Blue and locations 2-15 are Red
In a multiset, like in a set, the order of elements does not matter, so 14 reds and 1 blue would be a single multiset.

6. ## Re: How many unique combinations in this scenario?

Why don't you give us a complete example with, say, using 3 out of 4 colors.
Show ALL possibilities...shouldn't take long...

7. ## Re: How many unique combinations in this scenario?

Originally Posted by Plato
I find it hard to follow that example.
Is the arrangement of beads 1-9 & 11-15 being red and bead 10 blue the same as the above?.
Yes, exactly. Instead of beads, my real-life problem involves light bulbs. In both examples above, the total blended color of light in a sealed room would be the same. As long as there are 14 reds and 1 blue.

Hope this helps!

8. ## Re: How many unique combinations in this scenario?

Hi Wilmer. See my last post, and let me know if this clarifies the question.

9. ## Re: How many unique combinations in this scenario?

Originally Posted by MathinTExas
Yes, exactly. Instead of beads, my real-life problem involves light bulbs. In both examples above, the total blended color of light in a sealed room would be the same. As long as there are 14 reds and 1 blue.
Then reply #2 is correct. These are also known as multi-selections.
$\displaystyle \binom{N+k-1}{N}$ select $\displaystyle N$ items from $\displaystyle k$ different types.
There twenty kinds of doughnuts, how many ways can we select a dozen :
$\displaystyle \binom{12+20-1}{12}$.

10. ## Re: How many unique combinations in this scenario?

Originally Posted by Plato
Then reply #2 is correct. These are also known as multi-selections.
$\displaystyle \binom{N+k-1}{N}$ select $\displaystyle N$ items from $\displaystyle k$ different types.
There twenty kinds of doughnuts, how many ways can we select a dozen :
$\displaystyle \binom{12+20-1}{12}$.
Great. Please excuse my lack of experience with math symbols, and let me know the final number from this equation. Very interested in knowing this number based on choosing 15 selections from the pool of 16.

11. ## Re: How many unique combinations in this scenario?

$\displaystyle \left(\!\!{16\choose 15}\!\!\right)={16+15-1\choose 15}={30\choose15}=\frac{30!}{15!\cdot15!}$ where $\displaystyle n!=1\cdot2\cdot\ldots\cdot n$. My calculations show that this number is 155,117,520.

12. ## Re: How many unique combinations in this scenario?

Originally Posted by MathinTExas
Great. Please excuse my lack of experience with math symbols, and let me know the final number from this equation. Very interested in knowing this number based on choosing 15 selections from the pool of 16.
$\displaystyle \binom{15+16-1}{15}=\binom{30}{15}=\frac{30!}{(15!)(15!)}=15511 7520$

13. ## Re: How many unique combinations in this scenario?

Originally Posted by emakarov
$\displaystyle \left(\!\!{16\choose 15}\!\!\right)={16+15-1\choose 15}={30\choose15}=\frac{30!}{15!\cdot15!}$ where $\displaystyle n!=1\cdot2\cdot\ldots\cdot n$. My calculations show that this number is 155,117,520.
THANK YOU!!!! What a great forum :-)