Question

ind the number of integer solutions to the equation x1+x2+x3=40 where, 6<=x1>=15, 5<=x2>=20 and 10<=x3>=25

Answer 1

Answer on Question #69951 – Math – Discrete Mathematics

Question

Find the number of integer solutions to the equation $x_{1} + x_{2} + x_{3} = 40$ where

$$6 \leq x _ {1} \leq 15, \quad 5 \leq x _ {2} \leq 20 \quad \text{and} \quad 10 \leq x _ {3} \leq 25.$$

Solution

Method 1

Let $x_{1}, x_{2}, \ldots, x_{m}$ be integers. Then the number of solutions to the equation

$$x _ {1} + x _ {2} + \cdots + x _ {m} = n$$

subject to the conditions $a_1 \leq x_1 \leq b_1$, $a_2 \leq x_2 \leq b_2$, ..., $a_m \leq x_m \leq b_m$

is equal to the coefficient of $x^n$ in

$$(x ^ {a _ {1}} + x ^ {a _ {1} + 1} + \cdots + x ^ {b _ {1}}) \cdot (x ^ {a _ {2}} + x ^ {a _ {2} + 1} + \cdots + x ^ {b _ {2}}) \cdots (x ^ {a _ {m}} + x ^ {a _ {m} + 1} + \cdots + x ^ {b _ {m}}).$$

(see http://www.careerbless.com/aptitude/qa/permutations_combinations_imp8.php)

Applying this theorem to this problem one gets that the number of solutions to the equation

$$x _ {1} + x _ {2} + x _ {3} = 40$$

subject to conditions

$$6 = a _ {1} \leq x _ {1} \leq b _ {1} = 15, \quad 5 = a _ {2} \leq x _ {2} \leq b _ {2} = 20, \quad \ldots, \quad 10 = a _ {3} \leq x _ {3} \leq b _ {3} = 25$$

is equal to the coefficient of $x^{40}$ in

$$\left(x ^ {6} + x ^ {7} + x ^ {8} + x ^ {9} + x ^ {10} + x ^ {11} + x ^ {12} + x ^ {13} + x ^ {14} + x ^ {15}\right).$$ $$\left(x ^ {5} + x ^ {6} + x ^ {7} + x ^ {8} + x ^ {9} + x ^ {10} + x ^ {11} + x ^ {12} + x ^ {13} + x ^ {14} + x ^ {15} + x ^ {16} + x ^ {17} + x ^ {18} + x ^ {19} + x ^ {20}\right).$$ $$\begin{array}{l} \left(x ^ {10} + x ^ {11} + x ^ {12} + x ^ {13} + x ^ {14} + x ^ {15} + x ^ {16} + x ^ {17} + x ^ {18} + x ^ {19} + x ^ {20} + x ^ {21} + x ^ {22} + x ^ {23} + x ^ {24} \right. \\ \left. + x ^ {25}\right) = \\ = x ^ {6} x ^ {5} x ^ {10} \left(1 + x + x ^ {2} + x ^ {3} + x ^ {4} + x ^ {5} + x ^ {6} + x ^ {7} + x ^ {8} + x ^ {9}\right). \end{array}$$ $$\cdot \left(1 + x + x ^ {2} + x ^ {3} + x ^ {4} + x ^ {5} + x ^ {6} + x ^ {7} + x ^ {8} + x ^ {9} + x ^ {10} + x ^ {11} + x ^ {12} + x ^ {13} + x ^ {14} + x ^ {15}\right).$$ $$\begin{array}{l} \cdot \left(1 + x + x ^ {2} + x ^ {3} + x ^ {4} + x ^ {5} + x ^ {6} + x ^ {7} + x ^ {8} + x ^ {9} + x ^ {10} + x ^ {11} + x ^ {12} + x ^ {13} + x ^ {14} + x ^ {15}\right) = \\ = x ^ {21} \left(1 + x + x ^ {2} + x ^ {3} + x ^ {4} + x ^ {5} + x ^ {6} + x ^ {7} + x ^ {8} + x ^ {9}\right). \end{array}$$ $$\cdot \left(1 + x + x ^ {2} + x ^ {3} + x ^ {4} + x ^ {5} + x ^ {6} + x ^ {7} + x ^ {8} + x ^ {9} + x ^ {10} + x ^ {11} + x ^ {12} + x ^ {13} + x ^ {14} + x ^ {15}\right).$$ $$\cdot \left(1 + x + x ^ {2} + x ^ {3} + x ^ {4} + x ^ {5} + x ^ {6} + x ^ {7} + x ^ {8} + x ^ {9} + x ^ {10} + x ^ {11} + x ^ {12} + x ^ {13} + x ^ {14} + x ^ {15}\right)$$

Thus, the number of integer solutions is 135.

Method 2

From the equation $x_{1} + x_{2} + x_{3} = 40$ we can express one of variables in terms of others: $x_{3} = 40 - (x_{1} + x_{2})$. Thus, $10 \leq 40 - (x_{1} + x_{2}) \leq 25$, or, $-30 \leq -(x_{1} + x_{2}) \leq -15$, or $15 \leq x_{1} + x_{2} \leq 30$.

For each integer from 15 to 30 let's count the number of pairs of integers $(x_{1}, x_{2})$ such that $6 \leq x_{1} \leq 15$, $5 \leq x_{2} \leq 20$.

Let $x_{1} + x_{2} = n$, where $15 \leq n \leq 30$. The minimal value of $x_{1}$ (even less than 6) for which there exists $x_{2}$ such that $x_{1} + x_{2} = n$ is $n - 20$, i.e. $n$ minus upper limit of $x_{2}$. Taking into account the lower limit of $x_{1}$ being 6, the minimal value of $x_{1}$ really is $\max \{(n - 20), 6\}$. Similarly, the maximal value of $x_{2}$ (even greater than 15) for which there exists appropriate $x_{2}$ is $n - 5$, i.e. $n$ minus lower limit of $x_{2}$. Taking into account the upper limit of $x_{1}$ being 15, the maximal value of $x_{1}$ really is $\min \{(n - 5), 15\}$. Then the number of pairs of integers $(x_{1}, x_{2})$ such that $6 \leq x_{1} \leq 15$, $5 \leq x_{2} \leq 20$ and $x_{1} + x_{2} = n$ is the number of integers between $a_{n} = \max \{(n - 20), 6\}$ and $b_{n} = \min \{(n - 5), 15\}$, i.e. $(b_{n} - a_{n} + 1)$. In the table below columns correspond to value of $n$ while blue cells in each column correspond to valid values of $x_{1}$:

For $15 \leq n \leq 26$, $a_{n} = 6$ and for the rest, $27 \leq n \leq 30$, $a_{n} = n - 20$. For $15 \leq n \leq 19$, $b_{n} = n - 5$, and for the rest, $20 \leq n \leq 30$, $b_{n} = 15$. Dividing $n$ into three groups: below 20 (A), above 26 (B) and between (C):

$$\begin{array}{l} A = \sum_{n=15}^{19} (b_n - a_n + 1) = \sum_{n=15}^{19} ((n - 5) - 6 + 1) = \sum_{n=15}^{19} (n - 10) = \sum_{n=5}^{9} n = (9 - 5 + 1) \frac{9 + 5}{2} \\ = 5 \times 7 = 35. \end{array}$$ $$B = \sum_{n=20}^{26} (b_n - a_n + 1) = \sum_{n=20}^{26} (15 - 6 + 1) = \sum_{n=20}^{26} 10 = 10(26 - 20 + 1) = 10 \times 7 = 70.$$ $$\begin{array}{l} C = \sum_{n=27}^{30} (b_n - a_n + 1) = \sum_{n=27}^{30} (15 - (n - 20) + 1) = \sum_{n=27}^{30} (36 - n) = \sum_{n=36-27}^{36-30} n = \sum_{n=9}^{6} n \\ = \sum_{n=6}^{9} n = (9 - 6 + 1) \frac{6 + 9}{2} = 4 \times 7.5 = 30. \end{array}$$ $$N = A + B + C = 35 + 70 + 30 = 135.$$

Answer: 135.

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