There are 24 chairs in a row on which we place 5 final test, with at most 1 test per chair. In how many ways can this be done if no two adjacent chairs can have an test?
using the method I got that there would be 19 stars and 5 bars which are tests, but I'm confused on what to do with the no two adjacent chairs can have an test part next
Best Answer
As tests are same, it is still a problem that can be solved using Stars and Bars method (or call them Sticks and Stones). First, we have $5$ chairs with tests. Now we are left with $19$ chairs. We place $4$ of them between the chairs with tests so chairs with tests are no longer adjacent to each another. Now we need to decide how many of the remaining $15$ chairs go in each of the $6$ spaces (between chairs with tests or at two ends). So we are looking for number of solutions to,
$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 15$, where $x_1, x_2, x_3, x_4, x_5$ and $x_6$ are non-negative integers.
Applying sticks and stones, the answer would be $ \displaystyle {15 + 6 - 1 \choose 6 - 1}$