$\{x_i\}_{i = 1}^7$ is a set of 7 integers that satisfy $1≤ x_i ≤ 8$. How many such ordered sets of $7$ integers are there, such that $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 - x_1x_2x_3x_4x_5x_6x_7 = 6?$
Ordered set of integers
0
$\begingroup$
combinatorics
algebra-precalculus
1 Answers
1
First sort them so that $y_i \le y_{i+1}$. We can worry about how many permutations exist for each group later. If $y_1\ge 2$ you are sunk, so $y_1=1$. Similarly for $y_2$, $y_3$, $y_4$, $y_5$, and even $y_6$ because products grow much faster than sums. Setting all these to $1$ gives $6+y_7-y_7=6$ so $y_7$ can be anything from $1$ through $8$. There is $1$ way to sort all $1$'s and there are $7$ ways to sort $6 \ \ 1$'s and one other number, for a total of $1+6\cdot 7=43$