Permutations with repetition
If each of \(r\) positions may be filled by any of \(n\) different objects and repetition is allowed, there are
\[n^r\]
Permutations of identical objects
Suppose \(r\) objects contain \(x_1\) identical objects of type 1, \(x_2\) of type 2, through \(x_k\) of type \(k\), where \(x_1+x_2+\cdots+x_k=r\).
\[\frac{r!}{x_1!x_2!\cdots x_k!}\]
The \(x_i!\) permutations of the identical objects of type \(i\) do not create distinct arrangements, so each repeated count is cancelled.
Specific objects combined
If \(k\) particular objects among \(r\) distinct objects must stay together, first arrange those \(k\) objects, then treat them as one entity with the remaining \(r-k\) objects.
\[k!(r-k+1)!\]
Specific objects separated
First arrange the other \(r-k\) objects. They form \(r-k+1\) gaps; choose \(k\) gaps and arrange the \(k\) specified objects in them.
\[(r-k)!\binom{r-k+1}{k}k!=(r-k)!{}^{r-k+1}P_k\]
Need help? Join our JC Math tuition classes.
Learn more