Number of combinations of marbles in a bag

[moonbase]

EDITED: Should be permutations not combinations

Homework Statement

You have a bag of 10 marbles. There are 4 red marbles, 3 yellow, 2 green, and 1 blue marble. You remove them from the bag one of the time without replacement. Assuming each color of marble is identical and it doesn't matter which specific marble of each color is chosen, calculate the number of possible permutations in which you can remove the 10 marbles.

Homework Equations

If the specific marble picked did matter, the answer would be 10! but I'm not sure how to apply the different odds of the colors.

The Attempt at a Solution

I found the number of possible colors that can be picked for each draw.

1st: 4
2nd: 3, 4
3rd: 3, 4
4th: 2, 3, 4
5th: 2, 3
6th: 2, 3
7th: 1, 2, 3
8th: 1, 2
9th: 1, 2
10th: 1

Then I multiplied every possible value: (4*4)*(3*6)*(2*6)*(1*4)=13824

Am I doing this wrong? And is there a more efficient way to figure it out?

 

[awkward]

Hi moonbase,

This is a poorly stated problem since it asks to "calculate the number of possible combinations in which you can remove the 10 marbles". In this sort of mathematics, "combinations" means an arrangement in which order does not matter. So if you believe "combinations" is being used in its proper sense, the answer is 1: the combination consisting of 4 reds, 3 yellows, 2 greens, and 1 blue. But I think that's not what is intended; I think the order is supposed to matter.

So assuming the order is significant, my suggestion is to start by asking how many arrangements there would be if all 10 marbles were considered different. Then maybe you can find a way to compensate for the colors.

You first try is not correct. It might be made to work, but's tricky because the decisions you make at one point (for example, is the first marble blue?) affect the number of choices later on, and it would take a lot of book-keeping to keep everything straight.

 

[moonbase]

Oh okay I put the problem up from memory but I messed up the wording a bit. Anyway, I already know 10! is what it would be if they were all considered different, but I really have no idea where to go from there. Though I did realize this:

10 ncr 1 = 10
10 ncr 2 = 45
10 ncr 3 = 120
10 ncr 4 = 210

The sum of these is 385 but I don't think that's the correct answer because the possibilities overlap everywhere. Any other tips?

 

[phinds]

This is a poorly stated problem since it asks to "calculate the number of possible combinations in which you can remove the 10 marbles". In this sort of mathematics, "combinations" means an arrangement in which order does not matter. So if you believe "combinations" is being used in its proper sense, the answer is 1: the combination consisting of 4 reds, 3 yellows, 2 greens, and 1 blue. But I think that's not what is intended; I think the order is supposed to matter.

Nicely stated !

 

[awkward]

Oh okay I put the problem up from memory but I messed up the wording a bit. Anyway, I already know 10! is what it would be if they were all considered different, but I really have no idea where to go from there. Though I did realize this:

10 ncr 1 = 10
10 ncr 2 = 45
10 ncr 3 = 120
10 ncr 4 = 210

The sum of these is 385 but I don't think that's the correct answer because the possibilities overlap everywhere. Any other tips?

OK, so the total number of arrangements is 10! if all the marbles are different. Now let's suppose that just two of the marbles are the same color, like the two green marbles, and all the others are different. Label the two green marbles $G_1$ and $G_2$. In half the 10! arrangements, $G_1$ and $G_2$ appear in that order; in the other half, they appear in the order $G_2$ and $G_1$. So we if consider $G_1$ and $G_2$ indistinguishable, there will be 10! / 2 total arrangements.

Think about what would happen if there were 3 green marbles, $G_1$, $G_2$ and $G_3$. How many orders could they appear in? What would this do to the original 10! arrangements if they were indistinguishable? If you figure this out, then you should be able to extend the method to more colors pretty easily.

 

[moonbase]

Okay so it would be 10!/(1!2!3!4!)=12600. Thanks!