There are two urns. Urm 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. 3
B. 36
C. 66
D. 108

Answer

To solve the problem of transferring balls between two urns, we can break it down into clear steps.

### Step-by-Step Solution:

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).

2. **Choose 2 Balls from Urn A:**
- We need to select 2 balls from the 3 red balls in Urn A. The number of ways to choose 2 balls from 3 is given by the combination formula \( \binom{n}{r} \), where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose.
- Therefore, the number of ways to choose 2 balls from 3 is:
\[ \binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3!}{2! \cdot 1!} = \frac{3 \times 2 \times 1}{(2 \times 1) \cdot (1)} = 3 \]

3. **Choose 2 Balls from Urn B:**
- Next, we choose 2 balls from the 9 blue balls in Urn B. The number of ways to choose 2 balls from 9 is:
\[ \binom{9}{2} = \frac{9!}{2!(9-2)!} = \frac{9!}{2! \cdot 7!} = \frac{9 \times 8 \times 7!}{(2 \times 1) \cdot 7!} = \frac{9 \times 8}{2} = 36 \]

4. **Calculate the Total Number of Ways:**
- Since the selections from each urn are independent, we multiply the number of ways to choose from Urn A and Urn B:
\[ \text{Total Ways} = \binom{3}{2} \times \binom{9}{2} = 3 \times 36 = 108 \]

### Final Answer:
The total number of ways in which the balls can be transferred between the two urns is **108**.

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