How many 7-digit even numbers less than 3 000 000 can be formed using all the digits1, 2, 2, 3, 5, 5, 6 ?

To solve this problem, we need to consider the conditions given:

1. The number must be a 7-digit number.
2. The number must be less than 3,000,000.
3. The number must be even.
4. We can only use the digits 1, 2, 2, 3, 5, 5, 6.

Let's solve this step by step:

Step 1: Since the number must be less than 3,000,000, the first digit can only be 1 or 2. So, we have 2 options for the first digit.

Step 2: The number must be even. This means the last digit must be 2 or 6. So, we have 2 options for the last digit.

Step 3: For the remaining 5 digits, we can use any of the 5 remaining digits (1, 2, 3, 5, 5). Since we have 2 digits repeating (2 and 5), we use the formula for permutations of multiset: n!/(r1!*r2!) where n is the total number of items, and r1 and r2 are the number of times each item repeats. So, we have 5!/(2!*2!) = 30 options for the remaining 5 digits.

Step 4: Multiply the number of options for each digit to get the total number of 7-digit even numbers less than 3,000,000 that can be formed using all the digits 1, 2, 2, 3, 5, 5, 6. So, we have 2 * 2 * 30 = 120 such numbers.