I've seen this Number of solutions to $x_1 +x_2 +x_3 +x_4 = 1097$ with multiple "at least" restrictions
and this Stars and bars (combinatorics) with multiple bounds.
But I am having trouble answering this problem
Two chess players, A and B, are going to play 7 games. Each game has three possible outcomes: a win for A (which is a loss for B), a draw (tie), and a loss for A (which is a win for B). A win is worth 1 point, a draw is worth .5 points, and a loss is worth 0 points.
How many possible outcomes for the individual games are there, such that A ends up with 4 points and B ends up with 3 points?
I know how to do this normally with cases and I get 357, which is correct. But I am interested in doing it a stars and bars way where the answer is some variation on $x_1 + x_2 + x_3 .. x_7 = 8$
Basically, we can find the # of ways for A to get 4 points by counting each star as a half point, so we have 8 stars to spread accross with 6 bars.
But each cell cannot hold more than 2 of the points.
The total number of solutions without restriction is ${ 14 \choose 6} = 3003$
But now we want it for $x_1 <=2 , x_2 <=2 … x_7 <= 2$
So I decided that I could calculate when one of the cells has 3 or more, and then through symmetry multiply by 7 and then through PIE get rid of duplicates.
So I decided
${ 14 \choose 6} – { 11 \choose 6} * 7 + { 10 \choose 6} * 7 … + { 6 \choose 6} * 7$ should give me the answer, as we remove all the cases that break our rule of only 2 stars per cell max. But this gives 805 as the answer which is wrong.
I think maybe now I must calculate $x_1 <= 2$ and $x_2 <= 2$ …, or for $x_1=0$ $x_1=1$ and $x_1 =3$ and the same for all 7 variables but then I'm basically just counting cases/doing it the original way I did it.
Best Answer
It should be $$ {14\choose6}-7{11\choose6}+{7\choose2}{8\choose6}=357 $$ ${7\choose2}$ because there are that many ways to choose two of the seven $x_i$ to exceed two; only three terms to compute because it's impossible for more than two of the $x_i$ to exceed two; and ${8\choose6}$ because I don't know where you got ${10\choose6}$ from.