How to mathematically calculate the indistinguisable and distinct of the following permutation problems?

Question

I'm having trouble calculating how many indistinguishable and distinct solutions there are for each problems. I'm pretty confident with some of my solutions, but could anyone show me mathematically why there's that many solutions. And I'm not too sure how many solutions there are for distinct 1 and 2.

1

5 colors- 2 purple, 2 red and 1 brown.

No same color are next each another.

Indistinguishable solutions - 12 solutions

Distinct solutions - 48 solutions

2

6 colors- 4 pink, 1 yellow and 1 black.

No more than 2 pinks in a row.

Indistinguisable - 8 solutions

Distinct - 192 solutions

3

8 colors- 2 blue, 2 gold, 3 yellow and 1 brown.

The colors from position 1 and 7 are different from each other.

The colors from positions 4 and 8 are the same.

Gold is never in position 1 nor 8.

Yellow is never in positions 2 and 3.

Yellow to the left of every blue.

Blue is never in position 6 and 7.

Indistinguisable solutions - 3 solutions

Distinct solutions - 72 solutions.

Answer 1

Tackle every problem with sets for first question consider these:

A: set in which the green balls are together

B: set in which the blue balls are together

A ^ B (A intersection B) : set in which both green balls are together and blue balls are also together

let N(X) represent no of entries in set X.

N(A) = 4!/2! (for indistinguishable case) 2 blue balls, 1 white ball, and a group of green balls. arrange these and nullify arranging 2 alike blue balls.
= 4!*2! (for distinguishable case) same arrangement as above but now arrange the distinct green balls as well. so, we have

                       indistinguishable case            distinguishable case 
       N(A) =            4!/2!                             4!*2!
       N(B) =            4!/2!                             4!*2!
       N(A^B) =          3!                                3!*2!*2!

so total different arrangements are:

answer = total solutions - (N(A) + N(B) - N(A^B))

indistinguishable case = 5!/(2!*2!) - (4!/2! + 4!/2! - 3!) = 30 -(12 + 12 - 6) = 12 distinguishable case = 5! - (4!*2! + 4!*2! - 3!*2!*2!) = 120 -(48 + 48 - 24) = 48

likewise you can approach other problems.