Breaking down combinations

"Permuting, then sorting, then removing duplicates."

Say we have 10 balls.

• 3 are red, 3 are blue, 4 are green.
• Label the balls R1, R2, R3, B1, B2, B3, G1, G2, G3, G4.

We can order them in 10! ways.

• Similarly, we can order the red balls in 3! ways, blue in 3!, and green in 4!.

Now say we don't care about the order of the red balls.

• i.e. we remove their labels - they're all just R now.
• Then we can order them in $\frac{10!}{3!}$ ways.

Now say we don't care about the order of the blue and green balls too.

• Remove their labels - they're all just B and G now.
• Then we can order them in $\frac{10!}{3!3!4!} = {10 \choose 3}{7 \choose 3}{4 \choose 4}$ ways.

??? idk about this ⇒ NONE OF THIS GETS RID OF ADJACENT DUPLICATES (E.G. {R1, R2} ≠ {R2, R1} WHEN THEY SHOULD BE EQUAL) SO IDFK

Now say we only want to pick 3, and we don't care about any of the labels.

• ?