How many solutions does the equation $n_1 + n_2 + n_3 + n_4 + n_5 = 20$ have in the positive integers if $n_1 < n_2 < n_3 < n_4 < n_5$? 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......
I have no idea how to proceed. Manually, I have done it
$$1+2+3+4+10$$
$$1+2+3+5+9$$
$$1+2+3+6+8$$
$$1+2+4+5+8$$
$$1+2+4+6+7$$
$$1+3+4+5+7$$
$$2+3+4+5+6$$
But is there any way I can do it by Permutation and Combination method?

Let $m_1 = n_1, m_2 = n_2 -1, m_3 = n_3 -2, m_4 = n_4 -3, m_5 = n_5-4$; then $m_1 \leq m_2 \leq \cdots \leq m_5$ and $m_1+m_2+m_3+m_4+m_5 = 10$. Thus we need the number of 5 partitions of 10, $P(10,5)$. Clearly, $P(10, 5) = 7$, using the recurrence $P(n,p) = P(n-1, p-1) + P(n-p,p)$.
