# COUNTING TECHNIQUES

In how many ways can we arrange 9 flowers when 4 are the same?

F1, F2, F3, F3, F4, F3, F5, F3, F6

lets make the same flowers distinct by denoting them as: F1, F2, F31, F32, F33, F34, F4, F5, and F6

We know that the **number of permutations** (7 arrangements among 7) is 7! (given by n!)

Now for our set {F1, F2, F3, F3, F4, F3, F5, F3, F6 }, we have 4! permutations of F3 that do not affect the place of the five remaining elements F1, F2, F4, F5 and F6. (you can check this by listing the possible arrangements of the set and counting those arrangements in which the place of all distinct elements remain the same : there will be 24 such cases)

Therefore, for each permutation of 9 elements that we have for the set {F1, F2, F3, F3, F4, F3, F5, F3, F6

}, 3! permutations are identical.

Then by the **multiplication principle** :

The number of permutations of 9 different elements =

(The number of permutations of 9 elements which contains 4 identical elements) and (the number of permutations of the 4 identical elements),

That is,

(the number of permutations of 9 elements which contains 4 identical elements) x (the number of permutations of the 4 identical elements), that is:

upon rearranging :

(the number of permutations of 9 elements which contains 4 identical elements) x 4! = 9!

And therefore the number of permutations of 9 elements which contains 4 identical elements is 9! / 4!

**Hence the general rule:
The number of permutations of n elements which contains r identical elements is given by n!/r!**