My apologies for the vague title, I truly couldn't think of a better one. The question itself is quite tame compared to what's usually asked here, but is as follows:
Suppose you have n balls and k different "boxes" to put them in. Both the boxes and the balls are said to be identical. I have found two ways to compute the number of items in the sample space, n(S) , however, they give different answers. They are:
Method one: Take a singular ball. The ball has k different "options" for which ball to go into (As each ball can go into any one of the boxes). Over n balls, this amounts to k * k * k... giving $k^n$ as the value for n(S)
(While I do see some flaws with this, namely that initially there is no way that there are k options as all the boxes are (at least initially) identical, this leads to the official answer in a JEE mains 2015 question, given at the end of the question. So could both methods be wrong ?)
Method two (using combinatorics): This is more hack-ey, but I can't find any logical errors:
Take the n balls, and arrange them in a list first: [b,b,...b]. As to split a set into k other sets, you need k-1 divisions/lines to seperate them, add that many lines in the list. These effectively divides the list into k pieces which can be thought each of as a box.
Now (for example) the list looks like [b,b,|,b,b....b], with n(b) = n and n(k) = k-1
Now, using combinatorics, we get the possible number of configurations possible as $$n(s)=\dfrac{(n+k-1)!}{n!*(k-1)!}$$ Which gives quite different values from $k^n!$
UPDATE: Method 3: (from the official cengage answer key): I do not understand this well, thefeore copying it word for word (This, unlike the previous two answers gives the answer to the entire question, which is
If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes contains exactly 3 balls
I can't separate out the creation of the sample space, hence the whole answer:
Since the 3 boxes are identical, the probability of selecting a box in which we are to put 3 balls is 1/3
p = 1/3; q = 1 - 1/3 =2/3
Probability that the selected box should contain exactly 3 balls is the same as probability of exactly 3 succesess in 12 trials ( putting the 12 balls in 3 boxes).
∴ Required probability = $\binom nk *(\frac 1 3)^3*(\frac 2 3)^9 = (\frac {55} 3)*(\frac 2 3)^{11}$
So which is correct ? What is wrong with the other ones ? (Note: This question is a step in solving this one, where they use method one.)
Another way of thinking of the problem: How many ways are there of expressing $n$ as the sum of $k$ non-negative integers? The probability question at the end becomes "Of the ways of expressing $12$ as the sum of $3$ non-negative integers, how many include $3$ as one of the integers?"
The latter part of the last question has a straight-forward answer. If $3$ is one of the integers, then the sum of the other two must be $9$. As $9$ is not even, the two integers must be different. Thus there are $5$ choices for the lower integer: $0,1,2,3,4$. The upper integer is completely determined by this choice, so there are five ways of expressing $12$ as the sum of three non-negative integers, one of which is $3$.
To find the solution to the first question, develop a recursion: Define $S(n, k, m)$ to be the number of ways of expressing $n$ as the sum of $k$ integers $\ge m$. But if we have such a sum, we can subtract $m$ from each of the summands to get a sum equal to $n-km$ of non-negative integers. Thus $S(n,k,m) = S(n-km,k,0)$.
Now in a sum of $S(n,k,0)$, the smallest summand can be no greater than $\left\lfloor\frac nk\right\rfloor$, since if all the summands were higher than this, the summation would be greater than $n$. For each $0 \le m \le\left\lfloor\frac nk\right\rfloor$, the remaining $k-1$ integers must be $\ge m$ and sum to $n-m$. That is, $$\begin{align}S(n,k,0) &= \sum_{m=0}^{\lfloor n/k \rfloor}S(n-m,k-1,m) \\&= \sum_{m=0}^{\lfloor n/k \rfloor}S(n-n-(k-1)m, k-1,0) \\&= \sum_{m=0}^{\lfloor n/k \rfloor}S(n-km,k-1,0)\end{align}$$
Having served its purpose, drop the third argument by defining $S(n,k) = S(n,k,0)$, so $$S(n,k) = \sum_{m=0}^{\lfloor n/k \rfloor}S(n-km, k-1)$$
Note that $$S(n,2) = 1 + \left\lfloor \frac n2\right\rfloor$$ as the lower summand is $\le \left\lfloor \frac n2\right\rfloor$.
This is sufficient for calculating $S(12,3)$: $$\begin{align}S(12,3) &= S(12,2) + S(9,2) + S(6,2) + S(3,2) + S(0,2)\\ &=7 + 5 + 4 + 2 + 1\\&=19\end{align}$$
Thus I claim that the probability of putting twelve identical balls in three identical bins and having $3$ balls in (at least) one of the bins is $$\frac{S(9,2)}{S(12,3)} = \frac 5{19}$$
What went wrong with their calculation? What they calculated was the probability of picking a bin, and having that bin contain exactly $3$ balls, which while close, is not exactly the same thing.