The number of integers, between 100 and 1000 having the sum of their digits equals to 14, is _______

Q: Why can't we just use the standard stars and bars formula $\binom{n+k-1}{k-1}$?

The standard stars and bars formula $\binom{n+k-1}{k-1}$ assumes that each variable can take any non-negative integer value without an upper bound. In our problem, each digit is restricted to be at most 9 (and the first digit $a$ is restricted to be at most 9, which becomes $a' \le 8$). Because of these upper bounds, the simple formula overcounts, so we must use generating functions or the principle of inclusion-exclusion to subtract the invalid cases.

Q: Why did we substitute $a = a' + 1$?

The constraint for the hundreds digit $a$ is $1 \le a \le 9$. The standard generating function method works best when variables start from zero. By setting $a = a' + 1$, we shift the range of $a$ to $0 \le a' \le 8$, which aligns perfectly with the other variables $b$ and $c$ and allows us to use the standard polynomial form $(1 + x + \dots + x^n)$.

Q: How do we handle the $(1-x)^{-3}$ term when expanding?

The expression $(1-x)^{-3}$ is a classic power series expansion. Using the generalized binomial theorem, $(1-x)^{-n} = \sum_{k=0}^{\infty} \binom{k+n-1}{n-1} x^k$. For $n=3$, the coefficient of $x^k$ is $\binom{k+3-1}{3-1} = \binom{k+2}{2}$. This allows us to easily find the coefficient of $x^{13}$ for any term in the numerator expansion.

Question

The number of integers, between 100 and 1000 having the sum of their digits equals to 14, is _________

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with finding the number of integers between 100 and 1000 whose digits sum to 14. Let the three-digit integer be represented as $abc$.

The range 100 to 1000 implies that $a$ is the hundreds digit, $b$ is the tens digit, and $c$ is the units digit. The condition is defined by the equation:
$$a + b + c = 14$$

The constraints are the heart of the problem. Since $a$ is the leading digit, it cannot be zero, so $1 \le a \le 9$. The other digits, $b$ and $c$, range from $0 \le b \le 9$ and $0 \le c \le 9$.

### The Art of Substitution

To utilize generating functions effectively, we shift the variables to start from zero. Let $a = a' + 1$, where $a' \ge 0$.

Substituting this into our sum, we get $(a' + 1) + b + c = 14$, which simplifies to:
$$a' + b + c = 13$$

The new constraints are $0 \le a' \le 8$, $0 \le b \le 9$, and $0 \le c \le 9$. We are now looking for the number of non-negative integer solutions to this equation.

### The Generating Function Magic

The number of solutions is the coefficient of $x^{13}$ in the product of the polynomials representing the choices for each digit. For $a'$, the choices are $(1 + x + x^2 + \dots + x^8)$. For $b$ and $c$, the choices are $(1 + x + x^2 + \dots + x^9)$.

We seek the coefficient of $x^{13}$ in the expression:
$$P(x) = (1 + x + \dots + x^8)(1 + x + \dots + x^9)^2$$

Using the geometric series formula $\sum_{i=0}^{n-1} x^i = \frac{1-x^n}{1-x}$, we rewrite the expression as:
$$P(x) = \left( \frac{1-x^9}{1-x} \right) \left( \frac{1-x^{10}}{1-x} \right)^2 = (1-x^9)(1-x^{10})^2(1-x)^{-3}$$

### The Expansion and Final Victory

Expanding the numerator, we obtain:
$$(1 - x^9)(1 - 2x^{10} + x^{20}) = 1 - x^9 - 2x^{10} + 2x^{19} + x^{20} - x^{29}$$

Since we only require the coefficient of $x^{13}$, we ignore terms with powers greater than 13. This leaves us with the relevant part of the numerator: $(1 - x^9 - 2x^{10})$.

We multiply this by the expansion of $(1-x)^{-3}$, which is given by the series $\sum_{k=0}^{\infty} \binom{k+2}{2} x^k$. To find the coefficient of $x^{13}$, we calculate:
$$1 \cdot \binom{13+2}{2} - 1 \cdot \binom{13-9+2}{2} - 2 \cdot \binom{13-10+2}{2}$$

Calculating these binomial coefficients:
$$\binom{15}{2} = 105, \quad \binom{6}{2} = 15, \quad 2 \cdot \binom{5}{2} = 20$$

Finally, we perform the subtraction:
$$105 - 15 - 20 = 70$$

There are exactly **70** such integers.

Animated Solution for Mathematics - Permutations and Combinations: The number of integers, between 100 and 1000 having the sum of their digits equals to 14, is _________

Enter Numerical Value:

Visualized Solution

Identify the Range

Define Digit Constraints

Variable Substitution

New Constraints

Generating Function Setup

Simplify using GP Formula

Expand the Expression

Apply Binomial Coefficient Formula

Calculate Individual Terms

Final Result

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Art of Substitution

The Generating Function Magic

The Expansion and Final Victory

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