The number of ways in which we can select 5 numbers from the set of numbers {1, 2, 3, ..., 25) such that none of the selections includes four consecutive numbers is :

Options

A. 53109
B. 13350
C. 10035
D. none of these

Answer

To solve the problem of selecting 5 numbers from the set {1, 2, 3, ..., 25} such that none of the selections includes four consecutive numbers, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**: We need to select 5 numbers from a total of 25, ensuring that no four consecutive numbers are included in our selection.

2. **Total Combinations Without Restrictions**: First, we calculate the total number of ways to choose 5 numbers from 25 without any restrictions. This is given by the combination formula:
\[ \binom{25}{5} \]
We can calculate this as:
\[ \binom{25}{5} = \frac{25!}{5!(25-5)!} = \frac{25!}{5! \cdot 20!} = \frac{25 \times 24 \times 23 \times 22 \times 21}{5 \times 4 \times 3 \times 2 \times 1} = 53130 \]

3. **Counting Invalid Selections**: Next, we need to subtract the cases where at least one set of four consecutive numbers is included.

- If we have four consecutive numbers, we can denote them as \(x, x+1, x+2, x+3\). The smallest value for \(x\) can be 1 (for the numbers 1, 2, 3, 4) and the largest can be 22 (for the numbers 22, 23, 24, 25). Thus, \(x\) can take values from 1 to 22, which gives us 22 possible sets of four consecutive numbers.

4. **Choosing the Fifth Number**: After choosing four consecutive numbers, we need to select one more number from the remaining numbers. The remaining numbers will be:
- If we select \(x, x+1, x+2, x+3\), the remaining numbers will be from the set {1, 2, ..., 25} excluding \(x, x+1, x+2, x+3\). This means we have \(25 - 4 = 21\) numbers left to choose from.

5. **Calculating Invalid Combinations**: For each of the 22 sets of four consecutive numbers, we can choose the fifth number in 21 ways. Therefore, the total number of invalid selections is:
\[ 22 \times 21 = 462 \]

6. **Final Calculation**: Now, we subtract the invalid combinations from the total combinations:
\[ \text{Valid combinations} = \binom{25}{5} - \text{Invalid combinations} = 53130 - 462 = 52668 \]

Thus, the number of ways to select 5 numbers from the set {1, 2, 3, ..., 25} such that none of the selections includes four consecutive numbers is **52668**.