Permutations when some objects are alike I am trying to find the total number of signals that can be created from $3$ pink, $3$ white and $2$ black flags when arranged in a straight line. But, only $5$ flags are allowed in a signal.
I know how to find permutations in this situation when all are taken at a time i.e. a signal has $8$ flags, which would be $\frac{8!}{3! \cdot 3! \cdot 2!}$, but cannot see how to start solving a scenario where all are not taken at the same time.
Formula when all are taken at a time with some like objects
$$\frac{n!}{a! \cdot b! \cdot c! \cdot d!...}$$
Question
Is there a modified formula that will give permutations of $n$ objects taken $r$ at a time when some of these objects are alike and $r < n$?
 A: Let's generalize the problem slightly and find the number of signals with $r$ flags, say $a_r$.  Let $f(x)$ be the exponential generating function for $a_r$, i.e.
$$f(x) = \sum_{r=0}^{\infty} \frac{1}{r!} a_r x^r$$
Since there are 3 pink, 3 white, and 2 black flags,
$$f(x) = \left( 1 + x + \frac{1}{2!} x^2 + \frac{1}{3!} x^3 \right)^2  \left( 1 + x + \frac{1}{2!} x^2 \right)$$
Expanding this polynomial (I cheated and used a computer algebra system), we find
$$f(x) = 1+3 x+\frac{9 x^2}{2}+\frac{13 x^3}{3}+\frac{35 x^4}{12}+\frac{17 \
x^5}{12}+\frac{35 x^6}{72}+\frac{x^7}{9}+\frac{x^8}{72}$$
so the number of signals with 5 flags is 
$$5! \; \frac{17}{12} = 170$$
A: I was able to solve this problem but not sure if there is a better and shorter method. I used a $case$ approach to solving this by listing all possible groups of 5 flags using the original collection of 8 flags and then within each group of 5 flags we apply the formula for permutations of 5 objects at a time.
Is there a better and shorter approach to this problem?
Since there are 3 W(hite), 3 P(ink) and 2 B(lack) balls, so the cases are as below. We start with cases having the maximum of each type of flag and then decrease the maximum of each flag type by 1 till we reach 1 count for each flag type. In each case we get a group of 5 flags which we simply arrange 5 at a time using standard formula.
The sum of all ways i.e. arrangements of 5 flags is the sum of all cases, which is 170. This is also the answer given at the back of the book.
Case 1: 3W
3W+2P : 5!/(3!.2!) = 10 ways
3W+2B : 5!/(3!.2!) = 10 ways
3W+1P+1B : 5!/(3!.1!.1!) = 20 ways
Case 2: 3P
3P+2W : 5!/(3!.2!) = 10 ways
3P+2B : 5!/(3!.2!) = 10 ways
3P+1B+1W: 5!/(3!.1!.1!) = 20 ways
Case 3: 2B
2B+3W
2B+3P
2B+1P+2W : 5!/(2!.2!.1!) = 30 ways
2B+2P+1W : 5!/(2!.2!.1!) = 30 ways
Case 4: 1W
1W+3P+1B
1W+2P+2B
Case 5: 2W
2W+2P+1B:  5!/(2!.2!.1!) = 30 ways
2W+1P+2B
Case 6: 1W
1W+3P+1B
1W+2P+2B
Case 7: 2P
2P+1W+2B
2P+2W+1B
Case 8: 1P
1P+2W+2B
1P+3W+1B
Case 9: 1B
1B+2W+2P
1B+3W+1B
1B+1W+3B
