How many ways are there to put seven white and two black billiard balls in nine pockets? Some of the pockets may remain empty and the pockets are considered indistinguishable.
Well, my solution goes like this:
We have $7$ white balls and $2$ black balls . We can distribute this $9$ pockes . So, we can represent this as: $x_1+....+x_9=9$ . Each of the integers $x_i$ denotes the number of balls in the ith packet(where $i=\{1,2,...,9\}$ ) . Now the number of solutions of the equation is $17\choose 8$ .
However, if we do it in the following method:
We first find the result for black and white balls. The number of ways of putting $7$ white balls in $9$ packets is $15\choose 8$ and the number of ways of putting $2$ black balls in $9$ packets is $10\choose 8$ . So, the total number of ways is $15\choose 8$$10\choose 8$.
However, both should give the same answer. Which method is valid? Why is the other one not valid? I am not getting it...
Personally, I regard the problem as complicated enough, and the numbers as small enough, and my knowledge of partitions non-existent enough, that my approach is to simply take off my shoes (and count).
I will divide up the situation into disjoint cases, form an enumeration for each case, and then add up all of the enumerations.
I will use the symbols $a-b-0 ~: ~a + b = 7$ to symbolize that $a$ white balls go in one pocket, $b$ white balls go in some other pocket, and none of the other pockets receive a white ball.
$\underline{\text{Preliminary Results}}$
First, I will list all of the possible (distinct) types of white ball distributions. Then, I will re-organize them into groups, where the enumeration for each element in a group is identical.
$a01: ~7-0.$
$a02: ~6-1.$
$a03: ~5-2.$
$a04: ~5-1-1.$
$a05: ~4-3.$
$a06: ~4-2-1.$
$a07: ~4-1-1-1.$
$a08: ~3-3-1.$
$a09: ~3-2-2.$
$a10: ~3-2-1-1.$
$a11: ~3-1-1-1-1.$
$a12: ~2-2-2-1.$
$a13: ~2-2-1-1-1.$
$a14: ~2-1-1-1-1-1.$
$a15: ~1-1-1-1-1-1-1.$
So, there are $15$ distinctive distributions of the white balls, and for each distribution, the number of distinct distributions of the black balls must be computed.
Since only $2$ black balls are to be distributed, the enumeration for each of the $15$ distributions listed above is not that bad. For a specific white-ball-distribution, the enumeration will depend on $2$ factors:
How many kinds of pockets are involved.
For example, in $a01$ there are only $2$ kinds, a $[7]$ and a $[0]$.
In contrast, in $a10$ there are $4$ kinds, a $[3],[2],[1],$ and $[0]$.
How many of the kinds of pockets are singletons.
For example, although $a02$ and $a04$ each have $3$ kinds of pockets, their enumerations will be different. This is because $a02$ has $2$ singletons, and $a04$ has only $1$ singleton.
I will divide up the distributions into cases.
I will let $T_k$ denote the enumeration for case $k$.
In general, within each case, there will be two subcases.
Either the $2$ black balls are together in the same pocket, or they are not.
I will use $T_{ka}$ to denote the enumeration where the $2$ black balls are together in the same pocket.
I will use $T_{kb}$ to denote the enumeration where the $2$ black balls are not together in the same pocket.
It is probably best to supply the formulas that will pertain, if a Case involves $r$ different types, where $s$ of the types are non-singletons. Here, it is assumed that $r,s \in \Bbb{Z^+}$, with $s \leq r$.
For $T_{ka}$, any one of the $r$ types might receive both balls into the same pocket. So, $T_{ka} = r.$
For $T_{kb}$, you start by pretending that you must select $2$ of the $r$ types, sampling without replacement. Then, since you are assuming that there are $s$ non-singletons, you have to increment by $s$ to accommodate the $s$ possibilities that the $2$ black balls might go into different pockets of the same type, where there are $s$ such (non-singleton) types.
So, $\displaystyle T_{kb} = \binom{r}{2} + s.$
So, the computation for Case $k$ will be
$$r + \binom{r}{2} + s.$$
$\underline{\text{Case 1 : 2 types, 1 of which is a singleton}}$
The only distribution that fits in this group is $a01$.
$$T_1 = 2 + \binom{2}{2} + 1 = 4.$$
$\underline{\text{Case 2 : 2 types, 0 singletons}}$
The only distribution that fits in this group is $a15$.
$$T_2 = 2 + \binom{2}{2} + 2 = 5.$$
$\underline{\text{Case 3 : 3 types, 2 singletons}}$
This includes $a02, a03,$ and $a05$.
$$T_3 = 3 + \binom{3}{2} + 1 = 7.$$
$\underline{\text{Case 4 : 3 types, 1 singleton}}$
This includes $a04, a07, a08, a09, a11, a12$ and $a14$.
$$T_4 = 3 + \binom{3}{2} + 2 = 8.$$
$\underline{\text{Case 5 : 3 types, 0 singleton}}$
This includes $a13$.
$$T_5 = 3 + \binom{3}{2} + 3 = 9.$$
$\underline{\text{Case 6 : 4 types, 3 singletons}}$
This includes $a06$.
$$T_6 = 4 + \binom{4}{2} + 1 = 11.$$
$\underline{\text{Case 7 : 4 types, 2 singletons}}$
This includes $a10$.
$$T_7 = 4 + \binom{4}{2} + 2 = 12.$$
Final Computation:
$$\sum_{k=1}^7 T_k \times f(k),$$
where $f(k)$ denotes how many of the $15$ distributions are represented by Case $k$.
$$(4 \times 1) + (5 \times 1) + (7 \times 3) + (8 \times 7)$$
$$+ ~(9 \times 1) + (11 \times 1) + (12 \times 1)$$
$$=~ 118.$$