gmatter0913
Thanks Zarrolou
I understood the solution, but I would need a little more help on some concepts.
PERMUTATIONS OF 'n' items having some identical elements:Number of permutations of 'n' items out of which p1 are alike, p2 are alike, p3 are alike and the rest are different is n!/p1!p2!p3!
Suppose, we have a set of 8 alphabets - {A, A, B, B, C, C, D, D}
The number of ways we can arrange the 8 alphabets is 8!/2!*2!*2!*2! (where we divide by 2! because we cannot distinguish between AA and its combination reversed, AA

)
So, just the way we have the above formula, what is the formula to get all the possible combinations of 3 elements in the above case.
Will it be? - 4c3*(2)*(2)*(2)/8c4
where 4c3 is choosing combination of 3 elements among 4 unique elements.
and (2)*(2)*(2) is for each of the 3 elements in the combination there are two options (two identical elements)
Is there a general formula used to determine combinations in cases such as above?The other doubt that I have is in the above set of alphabets {A, A, B, B, C, C, D, D}
how do I get the number of permutations of 3 elements? Is it 3!/{what?} because I don't know how many elements are repeated among those 3.I) I do not know a general formula for the above case (and probably there is none).
II)You cannot use n!/p1!p2!p3! for your second problem. You have to try to work with the other formulas you know.
This is how I personally would do it:
First I would consider the case in which all three elements are different: [A,B,C]-[B,C,D]-... and since here we are taking about permutations, the total number of cases is \(4*3*2=24\) (you can pick any letter for the first slot, three (four-the previous chosen) for the second slot and so on).
Then I would consider the case in which a pair of letters is repeated: [A,A,B]-[A,A,C]-...
You can pick the pair of identical letters in 4 ways, the remaining letter is 3 ways. But each combination can be arranged in 3 ways: example
A,A,B-A,B,A-B,A,A.
So the total number is \(4*3*3=36\). Overall the possible permutations are 60.