A set of 15 different words are given. In how many ways is it possible to choose a subset of not more than 5 words?

Options

A. `4944`
B. `4^(15)`
C. `15^4`
D. `4943

Answer

To solve the problem of choosing a subset of not more than 5 words from a set of 15 different words, we can use the concept of combinations.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to find the number of ways to choose subsets of words from a total of 15 words, where the size of the subset can be 0, 1, 2, 3, 4, or 5 words.

2. **Using Combinations**:
The number of ways to choose \( r \) items from \( n \) items is given by the combination formula:
\[ C(n, r) = \frac{n!}{r!(n-r)!} \]
In our case, \( n = 15 \) and \( r \) can be 0, 1, 2, 3, 4, or 5.

3. **Calculating Each Combination**:
We will calculate \( C(15, r) \) for \( r = 0, 1, 2, 3, 4, 5 \):
- \( C(15, 0) = 1 \) (Choosing no words)
- \( C(15, 1) = 15 \) (Choosing 1 word)
- \( C(15, 2) = \frac{15!}{2!(15-2)!} = \frac{15 \times 14}{2 \times 1} = 105 \) (Choosing 2 words)
- \( C(15, 3) = \frac{15!}{3!(15-3)!} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455 \) (Choosing 3 words)
- \( C(15, 4) = \frac{15!}{4!(15-4)!} = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 1365 \) (Choosing 4 words)
- \( C(15, 5) = \frac{15!}{5!(15-5)!} = \frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} = 3003 \) (Choosing 5 words)

4. **Summing the Combinations**:
Now, we add all these combinations together to find the total number of ways to choose a subset of not more than 5 words:
\[ \text{Total} = C(15, 0) + C(15, 1) + C(15, 2) + C(15, 3) + C(15, 4) + C(15, 5) \]
\[ \text{Total} = 1 + 15 + 105 + 455 + 1365 + 3003 = 4945 \]

5. **Final Answer**:
Therefore, the total number of ways to choose a subset of not more than 5 words from a set of 15 different words is **4945**.