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 (1) 36 (2) 66 (3) 108 (4) 3

Options

A. 36
B. 66
C. 108
D. 3

Explanation

To solve the problem, we need to determine the number of ways to select and transfer balls between two urns. Here's a step-by-step breakdown of the solution:

### Step 1: Identify the balls in each urn
- Urn A contains 3 distinct red balls: let's denote them as R1, R2, R3.
- Urn B contains 9 distinct blue balls: let's denote them as B1, B2, B3, B4, B5, B6, B7, B8, B9.

### Step 2: Calculate the number of ways to choose 2 balls from Urn A
- We need to select 2 balls from the 3 distinct red balls in Urn A.
- The number of ways to choose 2 balls from 3 is given by the combination formula \( C(n, r) = \frac{n!}{r!(n-r)!} \).
- Thus, the number of ways to choose 2 balls from Urn A is:
\[ C(3, 2) = \frac{3!}{2!(3-2)!} = \frac{3 \times 2 \times 1}{2 \times 1 \times 1} = 3 \]

### Step 3: Calculate the number of ways to choose 2 balls from Urn B
- Now we need to select 2 balls from the 9 distinct blue balls in Urn B.
- The number of ways to choose 2 balls from 9 is:
\[ C(9, 2) = \frac{9!}{2!(9-2)!} = \frac{9 \times 8}{2 \times 1} = 36 \]

### Step 4: Calculate the total number of ways to transfer the balls
- Since the selections from each urn are independent, we multiply the number of ways to select from Urn A by the number of ways to select from Urn B:
\[ \text{Total ways} = C(3, 2) \times C(9, 2) = 3 \times 36 = 108 \]

### Conclusion
The total number of ways to select and transfer the balls between the two urns is 108.

### Final Answer
Thus, the answer is (3) 108.

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