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

Imagine you are standing in a grand hall, holding ten distinct, numbered balls. Before you sit four distinct, labeled boxes. Your task is to distribute these balls randomly.

For each of the ten balls, you have four choices of boxes. The first ball can go into any of the four, and the second ball, independent of the first, also has four choices. This continues until the tenth ball.

By the fundamental counting principle, the total number of ways to distribute these ten distinct balls into four distinct boxes is $4\times 4\times \cdots \times 4$ (ten times). This gives us a total sample space of:

This is our denominator, the bedrock upon which our probability calculation rests.

Selecting the Special Boxes

Now, we narrow our focus to the specific event: two boxes must contain exactly two and three balls, respectively. We must first select two boxes out of the four, which is represented by ${}^4{C}_2$.

Once we have selected these two boxes, we must decide which one receives two balls and which one receives three. Since the boxes are distinct, the order matters, so we multiply by $2!$ to account for these two arrangements.

Thus, the number of ways to select and assign the counts to these boxes is:

This is the architectural foundation of our favorable outcomes.

Filling the Special Boxes

With the boxes chosen and their capacities defined, we now turn to the balls themselves. We have ten distinct balls and need to choose two of them to place into the first special box, which is done in ${}^{10}{C}_2$ ways.

With two balls placed, we are left with eight balls. From these eight, we must choose three to place into the second special box, which is done in ${}^8{C}_3$ ways.

The number of ways to fill these two boxes is the product of these combinations:

We are effectively partitioning our set of balls into specific groups.

The Final Distribution

We have placed five balls in total (two in one box, three in another). We have five balls remaining and two boxes remaining that have no specific constraints.

These five balls are free spirits; each of them can be placed into either of the two remaining boxes. Since each of the five balls has two choices, the number of ways to distribute them is:

This step is crucial—never forget the 'leftover' elements in a counting problem!

Calculating the Probability

Now, we bring it all together. The total number of favorable outcomes $n(E)$ is the product of all our independent steps:

Substituting the values, we have $12\times 45\times 56\times 32$. We want the probability $P(E)=\frac{n(E)}{n(S)}$.

Recall that $4^{10}=(2^2{)}^{10}=2^{20}$. Breaking down the numerator into powers of two:

Multiplying these, we get $2^2\times 3\times 2^3\times 7\times 2^5=2^{10}\times 21$. The numerator becomes $2^{10}\times 21\times 45=2^{10}\times 945$.

Finally, our probability is:

The complexity dissolves into elegance. This is the beauty of combinatorics—when you respect the structure of the problem, the math rewards you with simplicity.