After meeting with my supervisor

Still survive!!!
We already discussed about the possible combination of the 4 results from 4 methods.
He said what I wrote should be correct. Hooley!
So that means 15 combinations as I posted before.
The reason for this is the answer that I am not quite sure yet but I got some idea from a guy in Pantip (HotChoc). The following sentences is what he said:

"You can view the problem as putting letters (a,b,c,d) into slots (1,2,3,4).
The letter you pick first is not important, but the position is.

There are 4 'big' cases:

1. all 4 slots are occupied by the letter you pick. In this case, the number of results is 4C4 = 4!/(4-4)!4! = 1. 4C4 means from 4 slots, pick 4 slots to be occupied by the letter.
2. 3 slots are occupied by the letter picked. Number of results is 4C3 (its a combination because order is not important; the letter goes into slot 1,2,3 is the same as 2,3,1). 4C3 = 4!/(4-3)!1! = 4
3. 2 slots are occupied by the letter picked. Number of results is 4C2 = 6. the remaining 2 slots can have 2 more cases...both slots occupied by the same letter: 2C2 = 1 and 1 slot occupied by one letter 2C1 = 2. so total results is 6 * (2+1) = 18.
   But that is wrong. because the case where 1 of the remain two slots occupied by 1 letter is duplicated: aabc is the same as aacb. so instead of 2C1 = 2, we must somehow deduct the duplicate and make it = 1.
4. Another problem is that the case where the remain 2 slots occupied by 1 letter...
   this is in fact the mirror image of the big case so there are more duplicates.
   axay and xaya will be the same if x = y.

sorry i cannot think of an elegant solution right now. hopefully someone else can help or you can get some idea to continue it :)p.s. another option is to look at the problem as comparison like my post #7. you have 2^6 results. but of those results...there are some that will never happen (such as a=b, b=c a!=c). you have to deduct those out from 2^6."

Thanks for his comment!