# How many different arrangements are possible?

Ben wants to make a bracelet with the 9 marbles. How many different arrangements are possible? There are 3 red, 2 blue, and 4 green marbles.

For instance, red1, red2, red3, blue1, blue2, green1, green2, green3, green4 and blue1, blue2, red1, red2, red3, green1, green2, green3, green4 are different options, while red2, red1, red3, blue1, blue2, green1, green2, green3, green4 and red1, red2, red3, blue1, blue2, green1, green2, green3, green4 are the same.

Could anyone give me the answer or some hints? Thank you very much!

• HINT You can think about it this way: First mix up the red and blue ones, and then mix into that the green ones – Bram28 Nov 8 '17 at 3:06
• HINT2: Another way to think about it is: Suppose all 9 marbles were different .. then how many arrangements can you have? Now divide that by the number of arrangements of the red ones amongst each other, and same for blue and green. – Bram28 Nov 8 '17 at 3:09
• @Bram28 Thanks man! I find a solution which is almost, if not exactly, the same as your hint 2. I just need to verify it later. Thank you very much^_^ – Danny Nov 8 '17 at 3:16

Based on your description, I assume the marbles are indistinguishable. There are $9!$ ways to arrange the 9 marbles in a line, then we must divide by the $4!$ ways to arrange the green marbles (i.e. $9!$ overcounts 24 times because it is counting $G1, G2, G3, G4$ as the same as $G2, G1, G3, G4$, etc.), $3!$ ways to arrange the red marbles, and $2!$ ways to arrange the blue marbles.
Assuming rotational and reflectional symmetry are not relevant, the answer would be $\frac{9!}{4!*3!*2!}$ $=$ $1260$.
But, if rotational and reflectional symmetry are accounted for, we must divide by 9 for rotational symmetry because moving a marble to the end (since it's a bracelet) is still the same arrangement, and divide by 2 because flipping the bracelet is still the same arrangement as well, for a final answer of $\frac{1260}{9*2}$ $=$ $70$.