Number of ways in which three distinct numbers can be selected between 1 and 20 both inclusive, whose sum is even is______.

Explanation

To solve the problem of finding the number of ways to select three distinct numbers between 1 and 20 (inclusive) such that their sum is even, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Numbers**: The numbers from 1 to 20 consist of 10 even numbers (2, 4, 6, ..., 20) and 10 odd numbers (1, 3, 5, ..., 19).

2. **Determine Cases for Even Sum**: The sum of three numbers can be even in two scenarios:
- Case 1: Select 3 even numbers.
- Case 2: Select 2 odd numbers and 1 even number.

3. **Calculate Case 1 (3 Even Numbers)**:
- The number of ways to choose 3 even numbers from 10 even numbers 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.
- Thus, the number of ways to select 3 even numbers is:
\[ \binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \]

4. **Calculate Case 2 (2 Odd Numbers and 1 Even Number)**:
- The number of ways to choose 2 odd numbers from 10 odd numbers is:
\[ \binom{10}{2} = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45 \]
- The number of ways to choose 1 even number from 10 even numbers is:
\[ \binom{10}{1} = 10 \]
- Therefore, the total number of ways for this case is:
\[ \binom{10}{2} \times \binom{10}{1} = 45 \times 10 = 450 \]

5. **Total Ways**: Now, we add the results from both cases to find the total number of ways to select the numbers such that their sum is even:
\[ \text{Total} = 120 + 450 = 570 \]

### Final Answer:
The total number of ways to select three distinct numbers between 1 and 20 such that their sum is even is **570**.