Understanding the Problem: Sum of Three-Digit Numbers

The question asks us to find the sum of all possible three-digit numbers that can be formed using the digits 3, 4, and 5, with the condition that no digit is repeated in any number. We need to use all three digits in each number.

Identifying Possible Three-Digit Numbers

Since we have three distinct digits (3, 4, and 5) and we need to form three-digit numbers using all of them without repetition, the possible numbers are the permutations of these three digits. The number of such permutations is given by $P(n, n) = n!$, where $n$ is the number of distinct digits. In this case, $n=3$.

Number of possible numbers = $3! = 3 \times 2 \times 1 = 6$.

The six possible three-digit numbers are:

• 345
• 354
• 435
• 453
• 534
• 543

Calculating the Sum of the Three-Digit Numbers

We can find the sum by adding these six numbers together.

Method 1: Direct Summation

Sum = $345 + 354 + 435 + 453 + 534 + 543$

Let's perform the addition:

$345$
$354$
$435$
$453$
$534$
$+ 543$
-----

Adding the units column: $5+4+5+3+4+3 = 24$. Write down 4, carry over 2.

Adding the tens column: $2 \text{ (carry)} + 4+5+3+5+3+4 = 26$. Write down 6, carry over 2.

Adding the hundreds column: $2 \text{ (carry)} + 3+3+4+4+5+5 = 26$. Write down 26.

Sum = 2664.

Method 2: Using Formula for Sum of Permutations

For $n$ distinct digits $d_1, d_2, \dots, d_n$, the sum of all $n$-digit numbers formed using these digits without repetition is given by the formula:

Sum = (Sum of the digits) $\times$ $(n-1)!$ $\times$ (Number formed by $n$ ones)

In this problem:

• The digits are 3, 4, and 5.
• Number of digits, $n=3$.
• Sum of the digits = $3 + 4 + 5 = 12$.
• $(n-1)! = (3-1)! = 2! = 2$.
• Number formed by $n$ ones (3 ones) = 111.

Using the formula:

Sum = $(3+4+5) \times (3-1)! \times 111$
Sum = $12 \times 2! \times 111$
Sum = $12 \times 2 \times 111$
Sum = $24 \times 111$
Sum = $2664$

Both methods give the same result. The sum of all three-digit numbers formed using the digits 3, 4, and 5 without repetition is 2664.

Conclusion

The sum of all three-digit numbers that can be formed using all the digits 3, 4, and 5 where repetition of digits is not allowed is 2664.

Revision Table: Key Concepts

Additional Information: Number Formation

When dealing with number formation problems using a given set of digits, it's important to consider whether repetition is allowed and whether all digits must be used.

• Repetition Allowed: If repetition is allowed, the number of possible numbers is calculated differently. For an $n$-digit number using $k$ distinct digits where repetition is allowed, there are $k$ choices for each position, so there are $k^n$ possible numbers.
• Repetition Not Allowed: As seen in this problem, if repetition is not allowed and all $n$ distinct digits must be used to form an $n$-digit number, the number of possibilities is $n!$ (permutations of $n$ distinct objects). If you need to form an $r$-digit number using $n$ distinct digits ($r \le n$) without repetition, the number of possibilities is $P(n, r) = \frac{n!}{(n-r)!}$.
• Digits Including Zero: If one of the digits is 0, special care must be taken for the first digit (hundreds place in this case) as it cannot be 0 for it to be a three-digit number.

The formula for the sum of permutations is particularly useful for competitive exams as it saves time compared to listing all numbers and adding them directly.