Animated Solution for Mathematics - Probability: If 10 different balls are to be placed in 4 distinct boxes at random, then the probability that two of these boxes contain exactly 2 and 3 balls is :

A Combinatorial Journey

My dear student, welcome to the beautiful world of combinatorics. Today, we are not just solving a probability problem; we are choreographing a dance of ten distinct balls into four distinct boxes.

It is a problem that tests not just your algebraic skills, but your ability to see the hidden structure within a set of conditions. Let us peel back the layers together.

The Universe of Possibilities

Before we can find the probability of our specific event, we must understand the entire universe of outcomes. Imagine you are holding ten distinct balls, each labeled from 1 to 10. You have four distinct boxes, labeled A, B, C, and D.

For the first ball, you have 4 choices. For the second ball, you also have 4 choices. This independence continues for all ten balls.

Thus, the total number of ways to distribute these balls is:

This is our sample space, the denominator of our probability fraction. It is a massive number, but do not be intimidated; we will handle it with elegance.

Decoding the Condition

Now, the heart of the problem: 'two of these boxes contain exactly 2 and 3 balls.' This phrasing is the trap. It does not mean only two boxes are involved.

It means that within our distribution, there must exist at least one box with 2 balls and at least one box with 3 balls. Let the counts of balls in the four boxes be $n_1,n_2,n_3,n_4$.

We know that $n_1+n_2+n_3+n_4=10$. We need the set $\{n_1,n_2,n_3,n_4\}$ to contain the values 2 and 3.

The Case of the Unique Counts

Let us first consider the scenario where we have exactly one box with 2 balls and exactly one box with 3 balls. The remaining two boxes must contain the rest of the balls, which is $10-2-3=5$.

Let these remaining counts be $x$ and $y$. We need $x+y=5$. Since we want to avoid creating extra boxes with 2 or 3 balls (which would lead us to our second case), $x$ and $y$ cannot be 2 or 3.

The possible pairs for $(x,y)$ are $(0,5)$ and $(1,4)$.

For the partition $(2,3,0,5)$, the number of ways to distribute the balls is given by the multinomial coefficient:

Similarly, for the partition $(2,3,1,4)$, the number of ways is:

Adding these together, we get $362880$ favorable outcomes for Case I.

The Case of the Double Counts

But wait! What if the remaining two boxes also contain 2 and 3 balls? This leads us to the partition $(2,2,3,3)$. This perfectly satisfies our condition of having a 2 and a 3.

To calculate this, we first choose which two boxes will receive 2 balls each: $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ ways. The remaining two boxes will automatically receive 3 balls each.

Then, we distribute the 10 balls into these fixed-capacity boxes:

Multiplying these, we get $6\times 25200=151200$ ways.

The Grand Finale

We have navigated the complexities. The total number of favorable outcomes is the sum of our cases: $362880+151200=514080$.

Now, for the final step, the probability $P$ is:

We can simplify this by writing $4^{10}$ as $2^{20}$. Dividing both the numerator and the denominator by $2^5=32$, we arrive at the elegant result:

See how the logic unfolds? It is not just about formulas; it is about partitioning the problem into manageable, logical stories. You have mastered the dance of the balls. Keep this clarity, and no JEE problem will ever be too daunting.