The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digit

Q: Why can't we just use the Stars and Bars formula $\binom{n+k-1}{k-1}$ directly?

Great question! The Stars and Bars formula $\binom{n+k-1}{k-1}$ counts non-negative integer solutions to $\sum x_i = n$ without any upper bound on the variables. Here, we have a strict upper bound: each digit $x_i$ cannot exceed $3$. Because our sum is small ($3$), we can handle this by partitioning, but for larger sums, we would need the Principle of Inclusion-Exclusion.

Q: How do I know I haven't missed a case?

By systematically partitioning the integer $3$. We are looking for sets of seven non-negative integers that sum to $3$. Since the maximum value any $y_i$ can take is $2$ (because $x_i \le 3$), we only have two ways to partition $3$: either $\{2, 1, 0, \dots\}$ or $\{1, 1, 1, 0, \dots\}$. Any other combination would violate the sum or the upper bound.

Q: Why did we use permutations of multisets?

Because the positions in our seven-digit number are distinct (the first digit is different from the second, etc.), but the values we are placing are identical. When we have a multiset like $\{2, 1, 0, 0, 0, 0, 0\}$, the number of ways to arrange them is the number of distinct permutations, which is given by $\frac{7!}{5!1!1!}$.

Question

The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only, is

Options:

(A) 55

(B) 66

(C) 77

(D) 88

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a combinatorics problem; we are learning how to dance with constraints.

Imagine you are standing before seven empty slots, waiting to be filled with the digits $1$, $2$, or $3$. You are told that the sum of these digits must be exactly $10$.

Let us represent our seven-digit number as $x_1, x_2, \dots, x_7$. We know that $x_i \in \{1, 2, 3\}$ and the constraint is:

$$ \sum_{i=1}^{7} x_i = 10 $$

The challenge is that the lower bound of $1$ makes the counting messy. So, let us simplify our reality.

### The Power of Variable Transformation

What if we shifted our perspective? Let us define a new variable $y_i = x_i - 1$.

Since the minimum value of $x_i$ is $1$, the minimum value of $y_i$ becomes $0$. This is a powerful normalization technique. If $x_i \in \{1, 2, 3\}$, then $y_i \in \{0, 1, 2\}$.

Now, let us substitute this into our sum equation:

$$ \sum_{i=1}^{7} (y_i + 1) = 10 $$

This expands to $\sum y_i + 7 = 10$, which simplifies beautifully to:

$$ \sum_{i=1}^{7} y_i = 3 $$

Suddenly, the problem is no longer about finding seven-digit numbers; it is about distributing a sum of $3$ into seven slots, where each slot can hold a value of $0, 1,$ or $2$. This is the elegance of algebra—turning a complex constraint into a manageable one.

### Partitioning the Sum

Since our target sum is only $3$, we do not need heavy machinery like generating functions. We can simply list the ways to partition $3$ using the allowed values $\{0, 1, 2\}$.

**Case 1:** We use one $2$ and one $1$. The remaining five slots must be $0$.

This gives us the multiset $\{2, 1, 0, 0, 0, 0, 0\}$. To find the number of ways to arrange these, we use the permutation formula for multisets:

$$ \frac{7!}{5! 	imes 1! 	imes 1!} = 42 $$

**Case 2:** We use three $1$s. The remaining four slots must be $0$.

This gives us the multiset $\{1, 1, 1, 0, 0, 0, 0\}$. Again, we calculate the permutations:

$$ \frac{7!}{4! 	imes 3!} = 35 $$

### The Final Synthesis

We have two mutually exclusive scenarios. By the fundamental principle of counting, we add them together:

$$ 42 + 35 = 77 $$

There are exactly **$77$** such integers.

Notice how we moved from a daunting constraint to a simple variable shift, and finally to a clear, logical partition. This is the essence of JEE Advanced problem-solving: visualize, transform, simplify, and conquer.

Animated Solution for Mathematics - Permutations and Combinations: The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only, is

Select Answer:

Visualized Solution

Visualizing the Seven-Digit Number

The Sum Equation

Variable Transformation: $y_{i} = x_{i} - 1$

The Transformed Sum Equation

Distributing a Sum of $3$

Case 1: One '2' and One '1'

Permutations for Case 1

Case 2: Three '1's

Permutations for Case 2

Total Number of Seven-Digit Integers

The Sigma Insight: Linear Permutations

Analyzing the Setup

The Power of Variable Transformation

Partitioning the Sum

The Final Synthesis

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