Counting 3-Digit Integers Whose Digit Sums Equal to 14
Counting the number of 3-digit integers between 100 and 1000 whose digits sum up to 14 is an interesting problem that involves combinatorial mathematics. This article will dive into the details of how to approach counting these integers, providing a comprehensive explanation along with relevant examples.
Introduction to the Problem
The problem is to find the count of 3-digit positive integers where the sum of the digits is equal to 14. To solve this, we need to explore permutations and combinatorial techniques. The solution will also involve understanding how to perform coefficient extraction from polynomial expansions.
Methodology
Given a 3-digit number (ABC), where (A), (B), and (C) are the hundreds, tens, and units digits, respectively, the sum of the digits is given by:
[ A B C 14 ]
Here, (A) ranges from 1 to 9 (since it is the hundreds place, it cannot be 0), and (B) and (C) range from 0 to 9. The problem can be approached by considering the coefficient of (x^{14}) in the expansion of the polynomial:
[ (x^0 x^1 cdots x^9)^3 ]
Polynomial Expansion
The polynomial can be simplified as:
[ left( frac{1 - x^{10}}{1 - x} right)^3 ]
This can be further decomposed into:
[ left( 1 - x^{10} right)^3 left( 1 - x right)^{-3} ]
Expanding further, we get:
[ left( 1 - 3x^{10} 3x^{20} - x^{30} right) left( 1 - 3x 6x^2 - 1^3 cdots right) ]
Considering the coefficients of (x^{14}) in the expansion, we find:
[ text{Coefficient of } x^{14} text{ in } (1 - 3x^{10} 3x^{20} - x^{30}) (1 - 3x 6x^2 - 1^3 cdots) ]
Removing the terms with powers higher than 14, we get:
[ text{Coefficient of } x^{14} text{ in } (1 - x)^{-3} - 3 times text{Coefficient of } x^4 text{ in } (1 - x)^{-3} ]
Using the binomial theorem, we have:
[ binom{14 3 - 1}{3 - 1} - 3 times binom{4 3 - 1}{3 - 1} ]
Which simplifies to:
[ binom{16}{2} - 3 times binom{6}{2} ]
Calculating the binomial coefficients:
[ 120 - 3 times 15 ]
[ 120 - 45 ]
[ 75 ]
Excluding Leading Zeros
Some of these 75 integers will have a leading zero if the sum of the tens and units digits is 14 or more. We need to exclude these cases. The count of 3-digit integers whose hundreds digit is zero and the sum of the remaining digits is 14 is simplified as:
[ text{Coefficient of } x^{14} text{ in } (x^0 x^1 cdots x^9)^2 ]
This can be simplified as:
[ binom{15}{1} - 2 times binom{5}{1} ]
Which simplifies to:
[ 15 - 2 times 5 ]
[ 15 - 10 ]
[ 5 ]
Final Count
Thus, the final count of 3-digit integers between 100 and 1000 whose digits sum up to 14 is:
[ 75 - 5 70 ]
Conclusion
Through a combination of combinatorial techniques and polynomial expansions, we have successfully determined that there are 70 such 3-digit integers. Understanding this problem not only enhances knowledge of combinatorial mathematics but also provides tools for solving similar problems in various contexts.