Find the number of three-digit numbers formed by using digits 1,2,3,4,6,7,8,9 without repetition such that sum of digits of the numbers formed is even.

Explanation

To solve the problem of finding the number of three-digit numbers formed using the digits 1, 2, 3, 4, 6, 7, 8, and 9 without repetition such that the sum of the digits is even, we can follow these steps:

### Step 1: Identify the digits
The available digits are 1, 2, 3, 4, 6, 7, 8, and 9.
- Odd digits: 1, 3, 7, 9 (Total: 4)
- Even digits: 2, 4, 6, 8 (Total: 4)

### Step 2: Determine conditions for an even sum
For the sum of three digits to be even, we have two cases:
1. All three digits are even.
2. Two digits are odd and one digit is even.

### Step 3: Case 1 - All three digits are even
We need to choose 3 even digits from the 4 available even digits:
- The number of ways to choose 3 even digits from 4 is given by \( \binom{4}{3} \).
- The arrangement of these 3 digits can be done in \( 3! \) ways.

Calculating this:
\[ \text{Ways} = \binom{4}{3} \times 3! = 4 \times 6 = 24 \]

### Step 4: Case 2 - Two odd digits and one even digit
We need to choose 2 odd digits from the 4 available odd digits and 1 even digit from the 4 available even digits:
- The number of ways to choose 2 odd digits from 4 is given by \( \binom{4}{2} \).
- The number of ways to choose 1 even digit from 4 is given by \( \binom{4}{1} \).
- The arrangement of these 3 digits can be done in \( 3! \) ways.

Calculating this:
\[ \text{Ways} = \binom{4}{2} \times \binom{4}{1} \times 3! = 6 \times 4 \times 6 = 144 \]

### Step 5: Total number of valid three-digit numbers
Now, we add the results from both cases:
\[ \text{Total} = \text{Case 1} + \text{Case 2} = 24 + 144 = 168 \]

### Final Answer
Thus, the total number of three-digit numbers that can be formed under the given conditions is **168**.

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