# Permutations with repetitions, when you're not picking all the objects?

If I have $6$ As, $9$ Bs and $6$ Cs, how many words/strings can I form with a length of 5?

I'm struggling a bit with this problem since now I'm restricted to picking only 5 letters in total. If I could pick all of them the solution would simply be

$$\frac{21!}{6! \times 9! \times 6!}$$

Is there a simple approach to such problem or do I have to count the number of ways of getting $1,2,3 ...$ As, and the number of ways to get $1,2,3 ...$ Bs, as well as the number of ways of getting $1,2,3 ...$ Cs and then subtract any overlapping strings?

• Can you please notify what the correct answer is, so that I could understand whether I have interpreted the question correctly or not. Dec 27 '17 at 17:38

In this case the number of strings is just $3^5$. Each letter of the string can be A, B or C, and you have more than enough of every letter, so you won't run out.
Often it is easier to do these problems by first calculating how many options there are if you ignore the restrictions, then subtracting off the ones that violate the restrictions. For example, if you wanted the number of strings of length $8$, there are $3^8$ strings in total, but $17$ have too many As and $17$ have too many Cs (in each case there are $16$ strings with exactly seven of that letter - you can choose the other letter in $2$ ways and its position in $8$ ways - and $1$ string with eight). So the answer would be $3^8-34$.
I actually didn't get your question correctly but what I understood was that how many words can you form with $6 A's$, $9 B's$ and$6 C's$. The answer is stated below.
1) All letters of the word are alike: $$^3C_1=3$$ 2) 4 letters alike and 1 letter different : $$^3C_1. ^2C_1. \frac{5!}{4!} =30$$ 3) 3 letters alike and 2 letters different: $$^3C_1. \frac{5!}{3!} =60$$
Summing up we get the answer as $93.$ You can comment and tell whether I have interpreted the question correctly or not.