I have this problem I am working on about permutations. I just want to know if what I am doing is correct.
A fair six-sided die is tossed four times and the numbers shown are recorded in a sequence. How many different sequences are there?
So by the rules of permutations, I used the formula:
$$P(n,r) = \frac{n!}{(n-r)!}$$
$$P(6,4) = \frac{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{2 \cdot 1}$$
which gave me the answer $360$. Is this correct? Another thought is saying since a dice has six sides, it only has six permutations.
Best Answer
No, this is not correct.
You are assuming that the digits cannot repeat, when in fact they can. There are $6$ possibilities for the first digit of the sequence, $6$ possibilities for the second, and $6$ possibilities for the third and fourth. The order of the digits matters.
Hence there are $6^4 = 1296$ different sequences.