I am trying to solve find the number of solutions to the inequality
$x_1 + x_2 + \dots + x_7 \leq 86$
Where $x_i$'s are $\textbf{odd}$ positive integers, and the order matters.
If we ignore the constraint that $x_i$'s have to be odd, we can introduce another variable $x_8$ such that we have an equality: $x_1 + x_2 + \dots + x_7 + x_8 = 86$. Now a solution to this equation corresponds to a way of choosing $86$ items from a a set with $8$ elements such that there are $x_1$ items of type one, $x_2$ items of type two, etc. Thus, the number of solutions is equivalent to the number of $86-$combinations with repetition allowed from a set with $8$ elements, which is given by:
$\binom{86+8-1}{86} = \binom{93}{86} = \binom{93}{7}$
However, I am confused about how I can now find the number of solutions if $x_i$'s are odd. I would appreciate any help on this!
They all have to be odd, so put 1 into each of the 7 variables.
That leaves $86-7 = 79$
Every variable needs to remain odd so you can only change them by amounts of 2. So you discard the extra 1. That leaves 78. $78/2 = 39$ groups of 2 to distribute amount 7 variables.
Now do the same trick of adding an 8th variable for the $\le$:
${{8+39-1} \choose 39} = {46 \choose 39} = {46 \choose 7}$