Let S be the set of all passwords which are six to eight characters long, where each character is ei

Q: Why do we subtract the cases with no numbers instead of just counting the ones with numbers?

Great question! If you try to count passwords with 'at least one number' directly, you would have to calculate cases with exactly one number, exactly two, and so on, up to all numbers. That is a massive amount of work. By using the complementary counting method—Total minus None—we simplify the problem into just two calculations per length. It is the most efficient path to the solution.

Q: Why is the total number of passwords $10^k$ and the alphabet-only passwords $5^k$?

Think about the choices for each character position. For a password of length $k$, each position can be any of the 5 alphabets or 5 numbers, giving us $5 + 5 = 10$ options per position. Since there are $k$ positions, we multiply 10 by itself $k$ times, resulting in $10^k$. If we exclude numbers, each position only has 5 options (the alphabets), so we get $5^k$.

Q: How do I know when to factor out $5^6$?

Always look at the target form given in the problem! The question asks for the answer in the form $\alpha \times 5^6$. This is a huge hint. It tells you that $5^6$ is a common factor you will eventually need to isolate. Once you group your terms, you will see $10^6$ and $5^6$ appearing; since $10^6 = (2 \times 5)^6 = 2^6 \times 5^6$, you can easily pull out that $5^6$.

Question

Let S be the set of all passwords which are six to eight characters long, where each character is either an alphabet from {A, B, C, D, E} or a number from {1, 2, 3, 4, 5} with the repetition of characters allowed. If the number of passwords in S whose at least one character is a number from {1, 2, 3, 4, 5} is $\alpha 	imes 5^6$, then $\alpha$ is equal to ______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to tackle a problem that might look like a tedious counting exercise, but beneath the surface, it is a beautiful lesson in strategic thinking.

We are tasked with finding the number of passwords of length six to eight that contain at least one number from the set $\{1, 2, 3, 4, 5\}$, given that our character set also includes five alphabets $\{A, B, C, D, E\}$.

### The Power of Complementary Counting

When you see the phrase 'at least one' in a combinatorics problem, your brain should immediately sound an alarm. Calculating 'at least one' directly is often a trap, as it forces you to sum up cases for one number, two numbers, three numbers, and so on.

Instead, we use the principle of **complementary counting**. We define our set of valid passwords $S$ as the total number of possible passwords minus the number of passwords that contain absolutely no numbers.

This simple shift in perspective turns a mountain of work into a molehill.

### Building the Mathematical Foundation

Let us define our character pool. We have 5 alphabets and 5 numbers, giving us a total of 10 distinct characters.

For any password of length $k$, the total number of unrestricted passwords is $10^k$. If we want to exclude all numbers, we are left with only the 5 alphabets.

Thus, the number of passwords with no numbers is $5^k$. For a password of length $k$, the number of valid passwords is given by:

$$N(k) = 10^k - 5^k$$

Our problem allows lengths of 6, 7, and 8. We calculate the valid passwords for each length:

For $k = 6$: $10^6 - 5^6$

For $k = 7$: $10^7 - 5^7$

For $k = 8$: $10^8 - 5^8$

### The Algebraic Symphony

Now, we sum these up to find the total $S$:

$$S = (10^6 - 5^6) + (10^7 - 5^7) + (10^8 - 5^8)$$

To make this manageable, we group the powers of 10 and the powers of 5:

$$S = (10^6 + 10^7 + 10^8) - (5^6 + 5^7 + 5^8)$$

Now, we factor out the smallest power from each group:

$$S = 10^6(1 + 10 + 10^2) - 5^6(1 + 5 + 5^2)$$

This simplifies to:

$$S = 10^6(111) - 5^6(31)$$

### The Final Reveal

The question asks for the answer in the form $\alpha 	imes 5^6$. We know that $10^6 = (2 	imes 5)^6 = 2^6 	imes 5^6$.

Substituting this back, we get:

$$S = (2^6 	imes 5^6)(111) - 5^6(31)$$

Now, we can factor out $5^6$:

$$S = 5^6 [ (2^6 	imes 111) - 31 ]$$

Since $2^6 = 64$, the expression becomes:

$$S = 5^6 [ (64 	imes 111) - 31 ]$$

Calculating $64 	imes 111$ gives us $7104$. Subtracting 31, we arrive at $7073$.

Thus, $S = 7073 	imes 5^6$. Comparing this to $\alpha 	imes 5^6$, we find that **$\alpha = 7073$**. You have just mastered the art of efficient counting!

Enter Numerical Value:

Visualized Solution

Identifying Available Characters

The 'At Least One' Strategy

Defining Password Lengths

Case 1: Length $k = 6$

Case 2 & 3: Length $k = 7$ and $k = 8$

Summing the Total Passwords

Factoring Out Common Terms

Relating $1 0^{6}$ to $5^{6}$

Final Arithmetic Calculation

Conclusion: Finding $α$

The Sigma Insight: Fundamental Principle of Counting

Analyzing the Setup

The Power of Complementary Counting

Building the Mathematical Foundation

The Algebraic Symphony

The Final Reveal

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