there are 3 pots and 4 coins .All these coins are to be distributed into these pots where any pot can contain any number of coins .
In how many ways all these coins can be distributed if all coins are different but all pots are identical ?

Options

A. a. 14
B. b. 21
C. c. 27
D. d. none of these

Answer

To solve the problem of distributing 4 different coins into 3 identical pots, we will consider the different distributions possible based on how the coins can be grouped.

### Step-by-Step Solution:

1. **Identify the Problem**: We have 4 different coins (let's label them as A, B, C, and D) and 3 identical pots. We need to find the number of ways to distribute these coins into the pots.

2. **Consider Different Cases**: Since the pots are identical, we can group the coins in various ways. We will consider the different distributions based on how many coins go into each pot.

3. **Case 1: All coins in one pot**:
- All 4 coins (A, B, C, D) go into one pot.
- There is only **1 way** to do this.

4. **Case 2: Three coins in one pot, one coin in another**:
- We can choose 3 coins to go into one pot and the remaining 1 coin will automatically go into another pot.
- The number of ways to choose 3 coins from 4 is given by \( \binom{4}{3} = 4 \).
- So, there are **4 ways** for this case.

5. **Case 3: Two coins in one pot, two coins in another**:
- We can choose 2 coins to go into one pot and the other 2 will go into another pot.
- The number of ways to choose 2 coins from 4 is \( \binom{4}{2} = 6 \).
- However, since the pots are identical, we have counted each distribution twice (e.g., {A, B} in one pot and {C, D} in another is the same as {C, D} in one pot and {A, B} in another).
- Therefore, we divide by 2 to avoid double counting: \( \frac{6}{2} = 3 \).
- So, there are **3 ways** for this case.

6. **Case 4: Two coins in one pot, one coin in another, and one coin in the last pot**:
- We can choose 2 coins to go into one pot, and the remaining 2 coins will each go into their own pots.
- The number of ways to choose 2 coins from 4 is \( \binom{4}{2} = 6 \).
- There is only **1 way** to distribute the remaining 2 coins into the other two pots (since each goes into its own pot).
- So, there are **6 ways** for this case.

7. **Total the Cases**:
- Now, we sum the number of ways from all cases:
- Case 1: 1 way
- Case 2: 4 ways
- Case 3: 3 ways
- Case 4: 6 ways
- Total = \( 1 + 4 + 3 + 6 = 14 \).

### Final Answer:
Thus, the total number of ways to distribute the 4 different coins into 3 identical pots is **14**.