Consider three more cases.

① Two pairs of two identical coloured balls (eg. two blue, two orange)

For the first pair, there are 5 colours to choose from→ 5 ways.
For the second pair, you have to choose 1 colour out of the remaining 4 colours→4 ways.

4 x 5 = 20 ways.

However, notice that you would have repeated some cases.
(eg. Orange Orange , Blue Blue is considered the same as Blue Blue, Orange Orange as your goal is to only select two pairs of identical coloured balls)

So we need to divide by 2! since permutation between pairs is not needed.

20 ÷ 2! = 10 ways

(Another way to calculate is : simply choose 2 colours out of the 5 for your two pairs. Then 5C2 = 10 ways)

② Only two balls are identical.

For this pair, there are 5 colours to choose from →5C1
= 5 ways
For the other two balls which must be of different colours, its just 4C2 (4 remaining colours ,choose 2 of them ) = 6 ways.

6 x 5 = 30 ways.


③ All 4 balls are of the same colour.

This one is easy. Simply just 5 colours to choose from → 5 ways.


Add the total number of ways here to your answers in (i) and (ii). They form all possible cases of selecting without restrictions.

5 + 20 + 10 + 30 + 5 = 70 ways.

thank you again!!