Using stars and bars counting strategy for De Mere's puzzle

Question

I am trying to solve this question (A fair die is rolled $3$ times. What is the probability that the sum of numbers in the $3$ throws is less than $6$?) and I was trying to calculate the number of events where the dice sum up to $11$ using number of non negative distinct integer solution technique.

where $6-x_i$ is the output of each die (as $5 \ge z_i$ and $1+z_i$ is the outcome of the die thrown so my new variable $z_i = 5 - x_i$ is )

using this technique my answer is

   6-x1 + 6-x2 +6-x3 = 11
=> 18 - (x1 + x2 + x3) = 11
=> 7 = x1 + x2 + x3
  

The number of non negative distinct integer solution is $(7+2)C2 = 36$ but the answer is $27$. Where am I going wrong?

Answer 1

Essentially you changed three usual dice rolls adding up to $11$ into three dice rolls adding up to $7$ where the new dice have values $0,1,2,3,4,5$ by subtracting each of the values of the old dice from $6$, and so the total from $18$.

You used stars and bars for the sum to $7$ to get the result ${7+2 \choose 2}=36$, but this assumes that $6$ or $7$ are possible values for the new dice. They are not, so you have to exclude:

• $7+0+0$ three times for the different permutations
• $6+1+0$ six times for the different permutations

which would give you $36-3-6=27$