Find the number of $7$-tuples $(x_{1},x_{2},x_{3}......,x_{7})$ such that all elements are distinct and $x_{i}∈ ℤ_{≥0}$.
Such that
$x_{1}+x_{2}+x_{3}+......+x_{7}=36$
My trial:
This is similar to the condition that $x_{i}<x_{i+1}$ with the only difference that I'll have to multiply my answer by $7!$
Then I assume:
$x_{1}=a_{1}$ and $x_{i+1}=x_{i}+1+a_{i+1}$,
where $a_{i}∈ ℤ_{≥0}$
Leaving me with:
$7a_{1}+6a_{2}+5a_{3}+.....+a_{7}=15$
Then I can't seem to proceed.
For example: ($3$-tuples)
$x_{1}+x_{2}+x_{3}=6$
the possible $3$-tuples are:
1)$(0,1,5),(0,5,1),(1,5,0),(1,0,5),(5,1,0),(5,0,1)$
2)$(0,2,4),(0,4,2),(2,4,0),(2,0,4),(4,2,0),(4,0,2)$
3)$(1,2,3),(1,3,2),(2,3,1),(2,1,3),(3,1,2),(3,2,1)$
the total $3-tuples$ are $(3)(3!)=18$
what I say is that assuming that $x_{i}<x_{i+1}$ gives me only three cases:
$(0,1,5),(0,2,4),(1,2,3)$
and finding this is sufficient for solving the question because the final answer would be achieved if I multiply this by $(3!)$. (i.e. answer$=3(3!)=18$)