Animated Solution for Mathematics - Permutations and Combinations: The number of 3-digit numbers, formed using the digits 2, 3, 4, 5 and 7, when the repetition of digits is not allowed, and which are not divisible by 3, is equal to__________

Mastering the Divisibility Constraint

Welcome, future engineers! Today, we are going to dive into a classic combinatorics problem that tests not just your ability to calculate, but your ability to strategize.

We are tasked with forming 3-digit numbers from the set $\{2,3,4,5,7\}$ without repetition, with the strict condition that these numbers must not be divisible by 3. Let's break this down into a journey of logical deduction.

The Total Universe

Before we worry about divisibility, let's look at the big picture. We have 5 distinct digits: $\{2,3,4,5,7\}$. We need to fill three slots: the Hundreds, the Tens, and the Units.

Since repetition is strictly forbidden, our choices diminish with each step. For the hundreds place, we have 5 options, for the tens place we have 4 options, and for the units place we have 3 options.

By the Fundamental Principle of Counting, the total number of 3-digit numbers we can form is:

This is our total universe of possibilities.

The Divisibility Trap

Now, we face the constraint: the numbers must not be divisible by 3. As we discussed in our strategy, counting the "not divisible" numbers directly is a nightmare.

Instead, we use the elegance of complementary counting. We will find the numbers that are divisible by 3 and subtract them from our total of 60.

We invoke the classic divisibility rule: a number is divisible by 3 if and only if the sum of its digits is a multiple of 3. Our mission is now clear: find all triplets from our set $\{2,3,4,5,7\}$ that sum to a multiple of 3.

The Systematic Hunt

Let's be methodical. We are looking for triplets whose sum is 9, 12, or 15. Let's test them:

1. Triplet 1: $\{2,3,4\}\Rightarrow 2+3+4=9$. This is a valid triplet!

2. Triplet 2: $\{2,3,7\}\Rightarrow 2+3+7=12$. This also works!

3. Triplet 3: $\{3,4,5\}\Rightarrow 3+4+5=12$. Another valid one!

4. Triplet 4: $\{3,5,7\}\Rightarrow 3+5+7=15$. Perfect!

If you test other combinations, like $\{2,4,5\}$ (sum 11) or $\{4,5,7\}$ (sum 16), you will see they fail the test. We have successfully identified exactly 4 valid triplets.

The Final Assembly

We aren't done yet! We have 4 sets of digits, but each set can be arranged in multiple ways to form distinct 3-digit numbers.

For any set of 3 distinct digits, the number of arrangements is given by:

So, for each of our 4 valid triplets, we can form 6 unique numbers. This gives us:

These are the numbers that are divisible by 3.

Finally, we return to our complementary counting strategy. We take our total universe of 60 numbers and subtract the 24 numbers that are divisible by 3.

The result is:

There you have it! By breaking the problem into manageable phases—calculating the total, identifying the constraint, hunting for valid triplets, and finally, performing the subtraction—we have navigated the complexity with ease. The final answer is 36.