Imagine the balls are of two colours but of different sizes. This makes each red ball and white ball unique.
The possible colour outcomes are RRR, RRW, RWR, RWW, WRR, WRW, WWR, WWW.
But when we consider each ball as unique we get different numbers. For example, RRR could be R1R2R3, R1R2R4, R1R2R5, etc. The number of permutations is 9*8*7=504. The number of combinations is 504/6=84, because the number of ways of arranging 3 balls is 6. The permutation R1R2R3 is not the same as R2R3R1, for example. But the combination R1R2R3 is the same as the combination R2R3R1.
(1) The balls are picked at random and returned after selection.
Now consider probabilities. P(R) is the probability of picking a red ball, so P(R)=9/15=3/5 and 1-P(R)=P(W)=2/5. There are 3 ways of picking exactly one red: RWW, WRW, WWR and so P(RWW)=P(WRW)=P(WWR)=(3/5)(2/5)^2=12/125. Add the three together and we get 36/125. That means that if the number of outcomes is N, then the number of ways of picking just one red is 36N/125. So we need to define and find N. Considering the uniqueness of each ball there are 15^3=3375 outcomes. So N=3375 and 36/125 of 3375=972. There are 972 ways of selecting exactly one red ball (any one of 9 reds) out of a possible 3375 outcomes.
(2) The three balls are selected at the same time without returning any.
This alters the probability and the number of outcomes. N=2730 (=15*14*13). The probabilities also change. Let's look at RWW, WRW, WWR. P(RWW)=9/15*6/14*5/13; P(WRW)=6/15*9/14*5/13; P(WWR)=6/15*5/14*9/13. Note that these all have the same value=9/91, so when added together we have 27/91. When we apply this to N=2730, we get 27*2730/91=810 ways.