Animated Solution for Mathematics - Permutations and Combinations: Total number of four digit odd numbers that can be formed using 0, 1, 2, 3, 5, 7 (using repetition allowed) are

Imagine you are standing in front of four empty boxes, each waiting to be filled by a digit from the set $\{0,1,2,3,5,7\}$. Your goal is to construct a $4$-digit odd number.

In the world of JEE Advanced, combinatorics is the art of counting without actually listing every possibility. Let us break this down systematically.

The VIP Seats (The Units Place)

Whenever you face a counting problem, your first instinct should be to identify the 'VIP' seats—the positions with the most restrictive rules. For an odd number, the VIP is the Units place.

A number is odd if and only if its last digit is odd. Looking at our available set $\{0,1,2,3,5,7\}$, the odd digits are $\{1,3,5,7\}$.

That gives us exactly $4$ choices for the Units place. We lock this in first to clear the path for the rest of the calculation.

The Thousands Trap

Now, let us look at the Thousands place. This is where many students stumble.

A $4$-digit number cannot start with a zero. If we allowed zero in the Thousands place, we would end up with numbers like $0135$, which is mathematically just $135$—a $3$-digit number.

Therefore, the Thousands place is also a restricted position. We must exclude $0$ from our set of $6$ digits, which leaves us with $\{1,2,3,5,7\}$, giving us $5$ valid choices for the Thousands place.

The Free Spirits (Hundreds and Tens)

With the VIP seats taken care of, we look at the Hundreds and Tens places. The problem states that repetition is allowed.

This is a massive relief! It means these positions have no restrictions.

We can use any of the $6$ digits—including zero—for both the Hundreds and the Tens places. So, we have $6$ choices for the Hundreds place and $6$ choices for the Tens place.

The Fundamental Principle of Counting

Now, we bring it all together using the Fundamental Principle of Counting. This principle states that if independent events occur in $m,n,p,q$ ways, the total combinations are the product of these ways.

Since we are filling all four boxes to form one single number, we multiply the number of choices for each box:

Substituting our values:

Let us calculate this smartly. Instead of multiplying left-to-right, group the numbers to make the math easier.

We know that $5\times 4=20$ and $6\times 6=36$. Now, the calculation becomes:

And there you have it! By respecting the constraints and applying the Fundamental Principle of Counting, we have determined that there are exactly $720$ such numbers.

Remember, the key to success in these problems is not speed, but systematic thinking. Always seat your VIPs first, and the rest will fall into place.