A shop sells 6 different flavors of ice-creams. In how many ways can a customer choose 4 ice-cream cones if they contain only 2 or 3 different flavors.

Options

A. a. 105
B. b. 462
C. c. 123
D. d. `""^(11)C_2`

Answer

To solve the problem of how many ways a customer can choose 4 ice-cream cones containing only 2 or 3 different flavors from 6 available flavors, we can break it down into two cases: one for 3 different flavors and another for 2 different flavors.

### Step-by-step Solution:

**Case 1: Choosing 3 Different Flavors**

1. **Choose 3 Flavors from 6**:
We need to choose 3 different flavors from the 6 available flavors. The number of ways to choose 3 flavors from 6 is given by the combination formula:
\[ \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]

2. **Distribution of Ice Creams**:
Since we have to choose 4 ice creams with 3 different flavors, one of the flavors must be chosen twice. Let's denote the chosen flavors as A, B, and C. The possible distributions of the ice creams can be:
- 2 of A, 1 of B, 1 of C
- 1 of A, 2 of B, 1 of C
- 1 of A, 1 of B, 2 of C

Each distribution can be arranged in different ways. The number of arrangements for each distribution can be calculated using the formula for permutations of multiset:
\[ \text{Arrangements} = \frac{4!}{2!1!1!} = 12 \]
(This is for the distribution 2 of A, 1 of B, and 1 of C. The same applies to the other distributions.)

3. **Total Arrangements for Case 1**:
Since there are 3 different distributions and each can be arranged in 12 ways, the total arrangements for this case is:
\[ 20 \times 12 = 240 \]

**Case 2: Choosing 2 Different Flavors**

1. **Choose 2 Flavors from 6**:
The number of ways to choose 2 different flavors from the 6 available flavors is:
\[ \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \]

2. **Distribution of Ice Creams**:
Since we have to choose 4 ice creams with 2 different flavors, each flavor must be chosen twice. Let's denote the chosen flavors as A and B. The only distribution is:
- 2 of A and 2 of B

The number of arrangements for this distribution is:
\[ \text{Arrangements} = \frac{4!}{2!2!} = 6 \]

3. **Total Arrangements for Case 2**:
The total arrangements for this case is:
\[ 15 \times 6 = 90 \]

**Final Calculation**:
Now, we combine the results from both cases to find the total number of ways a customer can choose the ice creams:
\[ \text{Total Ways} = 240 + 90 = 330 \]

### Final Answer:
The total number of ways a customer can choose 4 ice-cream cones containing only 2 or 3 different flavors is **330**.