Question Analysis:
We are given that N is a multiple of 72 and all the digits of N are different. We need to find the largest possible value of N.

Key Points:
- N is a multiple of 72.
- All the digits of N are different.

Solution:
To find the largest possible value of N, we need to arrange the digits in descending order.

Digits:
First, let's analyze the digits that can be used in N.

The largest digit is 9, as it will be the leftmost digit in N.
The next largest digit is 8, as it will be the second leftmost digit in N.
Continuing this pattern, the digits will be 9, 8, 7, 6, 5, 4, 3, 2, 1, 0.

Divisibility by 72:
To ensure that N is a multiple of 72, we need to check if it is divisible by both 8 and 9.

Divisibility by 8:
To check if a number is divisible by 8, we need to consider the last three digits. If the last three digits form a number divisible by 8, then the original number is also divisible by 8.

Divisibility by 9:
To check if a number is divisible by 9, we need to find the sum of its digits. If the sum of the digits is divisible by 9, then the original number is also divisible by 9.

Arranging the Digits:
Now that we have analyzed the digits and the divisibility conditions, let's arrange the digits in descending order to form the largest possible value of N.

Starting with the largest digit 9, we can place it in the leftmost position. Then, we continue with the next largest digit 8 and place it in the second leftmost position. We continue this pattern until we have placed all the digits.

The largest possible value of N:
The largest possible value of N, satisfying the conditions, is '9876543120'.

This number is divisible by 8 because the last three digits '120' form a number divisible by 8 (120/8 = 15).

This number is also divisible by 9 because the sum of its digits (9 + 8 + 7 + 6 + 5 + 4 + 3 + 1 + 2 + 0 = 45) is divisible by 9 (45/9 = 5).

Hence, '9876543120' is the largest possible value of N.