Q.Find the number of positive integers greater than $6000$ and less than $7000$ which are divisible by $5$, provided that no digit is to be repeated.

The key idea here is Permutations Without Repetition, as we are arranging distinct digits under specific constraints.

1. The integers are greater than $6000$ and less than $7000$, meaning they are 4-digit numbers of the form $6d_2d_3d_4$. Thus, the first digit is fixed as $6$.
2. For a number to be divisible by $5$, its last digit ($d_4$) must be $0$ or $5$.
3. We consider two cases based on the last digit, ensuring no digit is repeated:
   • Case 1: Last digit is $0$. The number is $6d_2d_30$. Digits $6$ and $0$ are used. The remaining two digits ($d_2, d_3$) must be chosen from the $8$ available digits $\{1, 2, 3, 4, 5, 7, 8, 9\}$ and arranged in $P(8,2)$ ways. This gives $8 \times 7 = 56$ numbers. …

To find the number of positive integers that meet all the given conditions, we will determine the number of choices for each digit position, moving from the most constrained positions first. The number must be a 4-digit number, say $d_1 d_2 d_3 d_4$, where $d_1$ is the thousands digit, $d_2$ is the hundreds digit, $d_3$ is the tens digit, and $d_4$ is the units digit.

Here are the conditions we must satisfy:

1. The number is greater than $6000$ and less than $7000$.
2. The number is divisible by $5$.
3. No digit is to be repeated.

Let's break this down step by step.

1. Determine the thousands digit ($d_1$).

   The condition that the number is greater than $6000$ and less than $7000$ immediately tells us that it must be a 4-digit number, and its thousands digit must be $6$.

   So, $d_1 = 6$.

   There is only $1$ choice for the thousands digit.

   Digits available for use: $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$.

   Digit used: $6$.

2. Determine the units digit ($d_4$).

   For a number to be divisible by $5$, its units digit must be either $0$ or $5$.

   So, $d_4 \in \{0, 5\}$.

   Since the thousands digit $d_1 = 6$ is neither $0$ nor $5$, there is no immediate conflict with the "no repeated digits" rule for $d_1$ and $d_4$. However, the choice of $d_4$ will affect the remaining available digits for $d_2$ and $d_3$. We will consider two separate cases based on the value of $d_4$.

3. Case 1: The units digit ($d_4$) is $0$.

   • Thousands digit $d_1 = 6$ (1 choice).
   • Units digit $d_4 = 0$ (1 choice).
   • Digits used so far: $6$ and $0$.
   • Remaining available digits for $d_2$ and $d_3$: From the set $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$, we exclude $6$ and $0$. This leaves us with $10 - 2 = 8$ digits: $\{1, 2, 3, 4, 5, 7, 8, 9\}$.
   • Determine the hundreds digit ($d_2$). We can choose any of the $8$ remaining available digits for $d_2$. So, there are $8$ choices for $d_2$.
   • Determine the tens digit ($d_3$). After choosing $d_2$, we have used three distinct digits ($d_1, d_4, d_2$). There are $10 - 3 = 7$ digits remaining for $d_3$. So, there are $7$ choices for $d_3$.
   • The number of integers in this case is $1 \times 8 \times 7 \times 1 = 56$.

4. Case 2: The units digit ($d_4$) is $5$.

   • Thousands digit $d_1 = 6$ (1 choice). …