Let $n_1 < n_2 < n_3 < n_4 < n_5$ be positive integers such that $n_1 + n_2 + n_3 + n_4 + n_5 = 20$. Then the number of such distinct arrangements $(n_1, n_2, n_3, n_4, n_5)$ is

Let n₁ < n₂ < n₃ < n₄ < n₅ be positive integers such that n₁ + n₂ + n₃ + n₄ + n₅ = 20. Then the number of such distinct arrangements (n₁, n₂, n₃, n₄, n₅) is

$\text{Given: } n_1 < n_2 < n_3 < n_4 < n_5,\quad n_i \in \mathbb{Z}^{+}$

$n_1 + n_2 + n_3 + n_4 + n_5 = 20$

$\text{To find: number of increasing tuples } (n_1, \dots, n_5)$

$\text{Plan: } 1.\ \text{subtract } 1, 2, 3, 4, 5$

$2.\ m_i = n_i - i \Rightarrow \text{non-decreasing}$

$3.\ \text{count the partitions of } 5$

$n_1 \ge 1,\ n_2 \ge 2,\ n_3 \ge 3,\ n_4 \ge 4,\ n_5 \ge 5$

$m_i = n_i - i \ge 0$

$n_1 < n_2 < \dots < n_5 \Rightarrow m_1 \le m_2 \le \dots \le m_5$

$\sum n_i = \sum (m_i + i) = \sum m_i + (1+2+3+4+5)$

$\Rightarrow 20 = \sum m_i + 15 \Rightarrow \sum m_i = 5$

$\text{Count non-decreasing } (m_1, \dots, m_5),\ m_i \ge 0,\ \sum m_i = 5$

$\text{i.e. count the partitions of } 5$

$5 = 5$

$5 = 4 + 1$

$5 = 3 + 2$

$5 = 3 + 1 + 1$

$5 = 2 + 2 + 1$

$5 = 2 + 1 + 1 + 1$

$5 = 1 + 1 + 1 + 1 + 1$

$\text{Total} = 7$

$(1,2,3,4,10),\ (1,2,3,5,9),\ (1,2,3,6,8)$ $(1,2,4,5,8),\ (1,2,4,6,7),\ (1,3,4,5,7)$ $(2,3,4,5,6)$

$\text{Number of arrangements} = 7$