0
$\begingroup$

As I understand there are $4$ cases for permutations:

$n$ objects, given by $n!$

$n$ objects, $k$ identical, given by $\frac{n!}{k!}$

$n$ objects, $r$ positions, given by $\frac{n!}{(n-r)!}$

$n$ distinct, $k$ identical, r positions, which is not in my textbook and I can't seem to be able to calculate : (

Please assist

For example lets assume we have objects AABC ($n=4,k=2$)

The possible arrangements for $r=2$ are:

  1. AA, AA
  2. AB, AB
  3. BA, BA
  4. AC, AC
  5. CA, CA
  6. BC
  7. CB

The right column refers to the same combination but switching the A with the second A. Since they are identical we don't care to count these as separate cases.

Thus in this case the answer to the 4th case should be $7$.

  • 0
    Is $S$ the same as $C$? – 2017-01-13
  • 0
    $AABC(n=4,k=2)$ then $n$ indicates the total of objects, and not the *distinct* ones (which should be $3$). And why do you consider $AA,AA$ as a possible "permutation" $r=2$ ? – 2017-01-13

1 Answers 1

0

As the problems become more complicated you have to derive the formula yourself because there are so many possibilities. Often it involves finding cases to break it into. Here I would consider the number of $A$s. You can have two $A$s in one way. If you don't have two $A$s, you are just choosing two ordered items from three, which is your third formula and evaluates to six. The total is then seven.