The total number of 3 -digit numbers, whose sum of digits is 10, is

Q: Wait, why can't we just use the Stars and Bars formula directly on $d_1 + d_2 + d_3 = 10$?

That is a fantastic question! The standard Stars and Bars formula $\binom{n+r-1}{r-1}$ counts the number of non-negative integer solutions (where variables can be 0). However, our first digit $d_1$ must be at least 1. If we used the formula directly, we would include cases where $d_1 = 0$, which would result in a 2-digit number (like $019$, which is just $19$). By substituting $d_1 = x_1 + 1$, we force $d_1$ to be at least 1, ensuring we only count valid 3-digit numbers.

Q: Why did we subtract 1 at the end? Is there a case where a digit becomes 10?

Exactly! The Stars and Bars method assumes each variable can take any non-negative integer value. In our transformed equation $x_1 + d_2 + d_3 = 9$, it is mathematically possible for $x_1$ to be 9 (making $d_1 = 10$), or for $d_2$ or $d_3$ to be 10. Since a digit must be $\le 9$, we must identify and remove these 'illegal' cases. We found that $d_1=10$ leads to the number $1000$, which is a 4-digit number, so we subtract that one invalid case.

Q: What if the sum of digits was 25? Would the subtraction method still work?

If the sum were larger, you might have more than one invalid case (e.g., $d_1=10, d_2=10$, etc.). In such scenarios, you would use the Principle of Inclusion-Exclusion. You would subtract cases where $d_1 \ge 10$, $d_2 \ge 10$, or $d_3 \ge 10$. It is a beautiful extension of the same logic, but for a sum of 10, our simple check is sufficient!

Question

The total number of 3 -digit numbers, whose sum of digits is 10, is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a vault, and to open it, you need to input a 3-digit number. The only clue you have is that the sum of its digits must be exactly 10.

We represent our 3-digit number as three empty boxes: $d_1, d_2, d_3$. The condition is defined by the equation:
$$d_1 + d_2 + d_3 = 10$$

However, the physical reality of a 3-digit number imposes a strict constraint: the first digit $d_1$ cannot be zero. Thus, our constraints are $1 \le d_1 \le 9$ and $0 \le d_2, d_3 \le 9$.

### The Art of Transformation

The Stars and Bars method is our most powerful tool for distributing a sum among variables, but it requires all variables to be non-negative (starting from 0). Our $d_1$ is a rebel; it starts from 1.

To tame it, we perform a simple algebraic shift. Let $d_1 = x_1 + 1$. Now, if $d_1 \ge 1$, then $x_1 \ge 0$.

Substituting this into our sum equation, we get:
$$(x_1 + 1) + d_2 + d_3 = 10$$

Subtracting 1 from both sides, we arrive at the elegant, normalized equation:
$$x_1 + d_2 + d_3 = 9$$

Now, every variable is free to be zero or greater.

### The Stars and Bars

We have 9 identical items (stars) to distribute among 3 distinct groups (variables). The formula for this is $\binom{n+r-1}{r-1}$, where $n=9$ and $r=3$.

This gives us:
$$\binom{9+3-1}{3-1} = \binom{11}{2}$$

Calculating this, we get:
$$\frac{11 	imes 10}{2 	imes 1} = 55$$

We have 55 potential candidates. But are they all valid?

### The Reality Check

We must ensure no digit exceeds 9. Our Stars and Bars calculation allowed variables to be 9 or more. Let us check the boundaries.

If $d_1 = 10$, then $x_1 = 9$, which forces $d_2 = 0$ and $d_3 = 0$. This corresponds to the number $1000$, which is a 4-digit number, not a 3-digit one. This is our one invalid case.

What if $d_2 = 10$? Then $d_1 + d_3 = 0$. Since $d_1 \ge 1$, this is impossible. The same logic holds for $d_3$.

Thus, we have exactly one invalid case to remove. Subtracting this from our total, we get:
$$55 - 1 = 54$$

There are exactly **54** such numbers. It is a triumph of logic over brute force, showing how a few simple transformations can unlock the secrets of combinatorics.

Animated Solution for Mathematics - Permutations and Combinations: The total number of 3 -digit numbers, whose sum of digits is 10, is

Enter Numerical Value:

Visualized Solution

Visualizing the 3-Digit Number

Defining the Constraints

Handling the Non-Zero Constraint

Simplifying the Equation

Applying Stars and Bars Method

The Stars and Bars Formula

Calculating the Combinations

Evaluating $(2 11)$

Checking the Upper Bound Constraints

Finding Invalid Cases

Verifying Other Digits

Final Calculation

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Art of Transformation

The Stars and Bars

The Reality Check

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