The number of five-digit numbers, greater than 40000 and divisible by 5, which can be formed using t

Q: Why can't we just multiply the number of choices for each slot? For $d_1$ we have $3$ choices, for $d_5$ we have $2$ choices, and for the middle $3$ slots we have $4 \times 3 \times 2$ choices. That gives $3 \times 2 \times 24 = 144$. Why is this wrong?

This is a classic trap! You assumed the choices for $d_1$ and $d_5$ are independent. But they are not. If you pick $5$ for $d_1$, you have only $1$ choice for $d_5$ (which is $0$). If you pick $7$ or $9$ for $d_1$, you have $2$ choices for $d_5$ ($0$ or $5$). The number of choices for $d_5$ depends on what you picked for $d_1$. When choices are dependent, you must use the Case Method.

Q: Why do we split the cases based on the first digit and not the last digit?

You could actually split by the last digit! If you fix $d_5 = 0$, then $d_1$ can be $5, 7, 9$ ($3$ choices). If you fix $d_5 = 5$, then $d_1$ can be $7, 9$ ($2$ choices). Both methods lead to the same result. The key is to pick the constraint that creates the most 'dependency' and resolve it first.

Q: What if the number was allowed to start with $0$?

If the number could start with $0$, the condition '$> 40,000$' would be violated unless the other digits were very large. But a $5$-digit number cannot start with $0$ by definition. If it did, it would effectively be a $4$-digit number. Always remember: the leading digit of an $n$-digit number cannot be $0$.

Question

The number of five-digit numbers, greater than 40000 and divisible by 5, which can be formed using the digits 0, 1, 3, 5, 7 and 9 without repetition, is equal to

Options:

(A) 132

(B) 120

(C) 72

(D) 96

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with forming a five-digit number using the set $S = \{0, 1, 3, 5, 7, 9\}$ under three specific constraints:
1. The number must be greater than $40,000$.
2. The number must be divisible by $5$.
3. No digit can be repeated.

Let the five slots be represented as $d_1, d_2, d_3, d_4, d_5$. The condition that the number is greater than $40,000$ implies that $d_1 \in \{5, 7, 9\}$. The condition that the number is divisible by $5$ implies that $d_5 \in \{0, 5\}$.

### The Tension of Constraints

The digit $5$ creates a dependency because it is a candidate for both the first slot ($d_1$) and the last slot ($d_5$). Because we cannot repeat digits, the choice made for $d_1$ directly restricts the available options for $d_5$.

To solve this, we must partition the problem into mutually exclusive cases based on the value of $d_1$.

### Case 1: The Digit $5$ at the Helm

Suppose $d_1 = 5$.

Since the digit $5$ is already occupied in the first position, it cannot be used in the last position. Therefore, $d_5$ must be $0$.

- $d_1$ has $1$ choice ($5$).
- $d_5$ has $1$ choice ($0$).
- The remaining $3$ slots ($d_2, d_3, d_4$) must be filled using the remaining $4$ digits from the set $S$.

The number of ways to arrange the middle digits is given by the permutation formula $P(4, 3)$:

$$4 	imes 3 	imes 2 = 24$$

Thus, Case $1$ yields **$24$** valid numbers.

### Case 2 & 3: The Digits $7$ and $9$ at the Helm

Now, consider the cases where $d_1 \in \{7, 9\}$. Let us analyze $d_1 = 7$ first.

In this scenario, $d_5$ can be either $0$ or $5$, as neither has been used yet. This provides $2$ choices for $d_5$.

- $d_1$ has $1$ choice ($7$).
- $d_5$ has $2$ choices ($0$ or $5$).
- The remaining $3$ slots are filled by the remaining $4$ digits.

The number of ways for $d_1 = 7$ is:

$$1 	imes 2 	imes (4 	imes 3 	imes 2) = 48$$

The logic for $d_1 = 9$ is identical to $d_1 = 7$. Consequently, $d_1 = 9$ also yields **$48$** valid numbers.

### Final Calculation

To find the total count of valid five-digit numbers, we sum the results of our mutually exclusive cases:

$$	ext{Total} = 24 + 48 + 48$$

$$	ext{Total} = 120$$

The total number of valid five-digit numbers is **$120$**.

Animated Solution for Mathematics - Permutations and Combinations: The number of five-digit numbers, greater than 40000 and divisible by 5, which can be formed using the digits 0, 1, 3, 5, 7 and 9 without repetition, is equal to

Select Answer:

Visualized Solution

Problem Setup

Analyzing Constraints

The Overlap Conflict

Case 1 Setup

Case 1 Execution

Case 2 Setup

Case 2 Execution

Case 3 Setup

Case 3 Execution

Final Summation

The Sigma Insight: Fundamental Principle of Counting

Analyzing the Setup

The Tension of Constraints

Case 1

Case 2 & 3

Final Calculation

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