There are $4$ balls of the 1st kind, $12$ balls of the 2nd kind and $3$ balls of the 3rd kind. There are $4$ boxes on a table. Balls of the 1st kind can be put only in 1st $2$ boxes and balls of 2nd kind can be put only in other $2$ boxes. Balls of 3rd kind can go in any box . In how many ways can the balls be put in these boxes? No two balls are same. Ordering of balls is not important.
I am able to place the 1st and 2nd kind of balls in respective boxes in $4C2 \cdot 12C2$ ways. But I'm not able to place 3rd kind of balls. How to include these balls? Proceeding with the most intuitive way creates a lot of cases which will not be effective if these numbers are large. Please suggest any effective method.
Thanks.
Since no two balls are the same, we have two choices for each of the four balls of the first kind. Hence, they can be distributed in $2^4$ ways. Similarly, we have two choices for each of the twelve balls of the second kind. Hence, they can be distributed in $2^{12}$ ways. However, we have four choices for each of the three balls of the third kind, so they can be distributed in $4^3$ ways. Thus, the number of ways of distributing the three types of balls given the restrictions is $$2^4 \cdot 2^{12} \cdot 4^3 = 2^4 \cdot 2^{12} \cdot 2^6 = 2^{22}$$
Note that $$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$ is the number of ways of selecting a subset of $k$ objects from a set with $n$ objects. Therefore, $$\binom{4}{2}$$ is the number of ways of selecting a subset of two of the four balls of the first kind, not the number of ways of distributing the balls of the first kind to two boxes. To verify that there are $2^4 = 16$ ways to distribute balls of the first kind to two boxes, we list the possibilities. Let's call the four distinct balls of the first kind $a, b, c, d$. Here are the possible distributions. $$ \begin{array}{c | c} \text{Box 1} & \text{Box 2}\\ \hline a, b, c, d & \\ a, b, c & d\\ a, b, d & c\\ a, c, d & b\\ b, c, d & a\\ a, b & c, d\\ a, c & b, d\\ a, d & b, c\\ b, c & a, d\\ b, d & a, c\\ c, d & a, b\\ a & b, c, d\\ b & a, c, d\\ c & a, b, d\\ d & a, b, c\\ & a, b, c, d \end{array} $$