There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each ur

Q: Wait, why do we use combinations ($^nC_r$) instead of permutations ($^nP_r$)?

That is a fantastic question! We use combinations because the problem asks for the number of ways to *select* the balls. In this scenario, picking ball $R_1$ then $R_2$ is the exact same outcome as picking $R_2$ then $R_1$. Since the order of selection within the pair does not change the set of balls being transferred, we use combinations to avoid overcounting.

Q: Why do we multiply the results instead of adding them?

Think of the Fundamental Principle of Counting. We are performing two independent tasks: selecting from Urn A AND selecting from Urn B. When you have to complete both tasks to finish the entire operation, you multiply the number of ways to perform each task. If we were choosing to perform *either* the selection from A *or* the selection from B, we would add them, but here, both must happen!

Q: What if the balls were identical? Would the answer change?

Absolutely! If the balls were identical, there would only be one way to select two balls from any urn, because all selections would look the same. The 'distinct' nature of the balls is what creates the variety in the selection process. Always check if your objects are distinct or identical before you start calculating!

Question

There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urn two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is

Options:

(A) 36

(B) 66

(C) 108

(D) 3

Sigma Percentile · Accepted Answer

### The Beauty of Combinatorial Selection

Welcome, fellow traveler on the journey of JEE Advanced preparation! Today, we are going to unravel a problem that might seem simple at first glance but holds a beautiful lesson in the Fundamental Principle of Counting.

Imagine you are standing before two urns, Urn A and Urn B. Urn A is filled with $3$ distinct red balls, and Urn B is filled with $9$ distinct blue balls. Our mission is to perform a simultaneous swap: we must select $2$ balls from Urn A to move to Urn B, and $2$ balls from Urn B to move to Urn A.

### Phase 1: The Red Ball Selection

First, let us focus our attention entirely on Urn A. We have $3$ distinct red balls and we need to choose $2$ of them.

Since the balls are distinct and the order in which we grab them does not matter, we use the combination formula:

$$^nC_r = \frac{n!}{r!(n-r)!}$$

Substituting $n=3$ and $r=2$, we calculate:

$$^3C_2 = \frac{3!}{2!(3-2)!} = \frac{3 	imes 2!}{2! 	imes 1!} = 3$$

There are exactly $3$ ways to choose our red balls.

### Phase 2: The Blue Ball Selection

Now, let us shift our gaze to Urn B. Here, we have $9$ distinct blue balls, and we must select $2$ of them to transfer to Urn A.

Using the combination formula for $^9C_2$, we substitute $n=9$ and $r=2$:

$$^9C_2 = \frac{9!}{2!(9-2)!} = \frac{9 	imes 8}{2 	imes 1} = 36$$

There are $36$ different pairs of blue balls we could potentially move.

### Phase 3: The Grand Synthesis

Now, we arrive at the most critical moment of our journey. We have calculated the ways to select from Urn A ($3$ ways) and the ways to select from Urn B ($36$ ways).

Because these two events are independent—the choice of red balls does not restrict our choice of blue balls—we invoke the **Fundamental Principle of Counting**. This principle states that if one task can be done in $m$ ways and another in $n$ ways, the total number of ways to perform both is $m 	imes n$.

Therefore, our total number of ways is:

$$3 	imes 36 = 108$$

### Conclusion

And there we have it! By breaking the problem into independent, manageable pieces, we have navigated the complexity and arrived at the final answer of **$108$**.

Remember, the key to mastering combinatorics is to always pause and ask: "Are these events independent?" and "Does the order of selection matter?" Keep practicing, keep visualizing, and most importantly, keep falling in love with the elegance of the math!

Animated Solution for Mathematics - Permutations and Combinations: There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urn two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is

Select Answer:

Visualized Solution

Visualizing the Setup

Selecting from Urn A

Calculating $^{3} C_{2}$

Selecting from Urn B

Calculating $^{9} C_{2}$

The Multiplication Principle

Final Computation

Summary and Takeaway

The Sigma Insight: Combinations and Selection

The Beauty of Combinatorial Selection

Phase 1

Phase 2

Phase 3

Conclusion

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