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?
[Math] How many solutions does the equation $n_1 + n_2 + n_3 + n_4 + n_5 = 20$ have in the positive integers if $n_1 < n_2 < n_3 < n_4 < n_5$
combinatorics
Best Answer
A variation based upon generating functions. We introduce positive integers $a,b,c,d$ and put \begin{align*} n_2&=n_1+a\\ n_3&=n_2+b=n_1+a+b\\ n_4&=n_3+c=n_1+a+b+c\\ n_5&=n_4+d=n_1+a+b+c+d \end{align*}
The equation $n_1+n_2+n_3+n_4+n_5=20$ transforms to \begin{align*} 5n_1+4a+3b+2c+d=20\tag{1} \end{align*} with $n_1,a,b,c,d>0$.
Add-on: Some details
We first transform the equation with restrictions by introducing positive integers $a,b,c,d$ in an equivalent equation with more convenient restrictions \begin{align*} &n_1 + n_2 + n_3 + n_4 + n_5 = 20\qquad&\qquad&5n_1+4a+3b+2c+d=20\\ &0<n_1<n_2<n_3<n_4<n_5\qquad&\qquad&0<n_1,0<a,0<b,0<c,0<d \end{align*}
We now consider admissible $5$-tuples $(n_1,a,b,c,d)$. Increasing $n_1$ by $1$ adds $5$ to the equation. Similarly, increasing $a$ by $1$ adds $4$ to the equation. We encode these increments via exponents of generating functions:
and similarly for $b,c$ and $d$. Observe that each of $n_1,a,b,c,d$ is positive, i.e. has at least value $1$. This is respected by smallest values $x^5,x^4,x^3,x^2$ and $x^1$.
Comment:
In (2) we use the coefficient of operator rule: $[x^{p}]x^qA(x)=[x^{p-q}]A(x)$.
In (3) we expand the geometric series restricted to powers less or equal to $x^5$ since other terms do not contribute to $[x^5]$.
In (4) we expand further and can omit terms with powers greater than $5$.
Hint: Instructive examples can be found in H.S. Wilf's book generatingfunctionology.