For 5 different elements and array with size 50 , all possible combinations is 5^50
Could you please tell me , what is the number of combinations when we include the condition that rotation does not make a new element?
I mean this combinations are the same: A,B,C,D=D,A,B,C=C,D,A,B
I am not at all sure that I follow your descriptions. That said, here is my take.
Lets suppose that is a five element set \(\{R,O,Y,G,B,V\}\) Five beads identical save for colour: Red, Orange, Yellow, Green , Blue, & Violet.
Trying to simplify somewhat, say that a string is made if fifteen of these beads. Lets also say that there are at least fifteen of each colour.
As you correctly note, there are \(5^{15}\) possible strings of length fifteen made up of selections from the given.
Your used the word
array which denotes
order. But if we deal with
circular arrangements the counting is different.
Some linear arrays would be \(RRRRRRRRRRRRRRR\) another \(RGRGRGRGRGRGRGR\) OR \(RRRRRRRRBBBBBBB\).
If we were dealing with five people there are \(5!=120)\) ways to form a queue where there are \((5-1)!=24\) ways to seat those around a circular table.
Suppose we select one R, two B's, three G's, four Y's, & five O's.
There are \(\dfrac{15!}{2!\cdot 3!\cdot 4!\cdot 5!}\) ways to form a queue.
But circular counts are more complicated.
