If 3 six faced fair dice are thrown together, then the probability that the sum of the numbers $ 9\leqslant k \leqslant 14$ If 3 six faced fair dice are thrown together, then the probability that the sum of the numbers appearing on the dice is k where $ 9\leqslant k \leqslant 14$ is  :
My approach : Possible sum values are from 3 to 18 and our cases of interst are $9,10,11,12,13,14 \therefore$ probability should be $\frac{6}{16}$ but the answer given is $\frac{21k-k^{2}-83}{216}$ How can arrive at the given answer ?
 A: Consider only Sum of all outcomes to be 9.
Then see the cases where when 3 fair dices are thrown, the sum comes out to be 9. There will be 25 such cases when checked.
Means probability will be 25/216.
With this knowledge, put k=9 in the given equations. The equation which will give the same output is the answer. Thus, equation with 21k-k^2-83 as numerator satisfies our requirement.
A: Every roll of three dice corresponds to an ordered triple $(D_1, D_2, D_3)$, where $1 \leq D_i \leq 6$ for each $1 \leq i \leq 3$. There are $6^3 = 216$ such ordered triples, which make up our sample space: $$\Omega := \{(D_1, D_2, D_3):\text{for each } 1 \leq i \leq 3, D_i \in \Bbb{N} \text{ and } 1 \leq D_i \leq 6 \} .$$ Therefore the probability we want to find, $$\Bbb{P}( D_1 + D_2 + D_3 = k ) \text{ for } k \in \Bbb{N} \text{ and } 3 \leq k \leq 18,$$ must have a denominator of $216$, if we write it as $$\Bbb{P}( D_1 + D_2 + D_3 = k ) = \frac{|E_k|}{|\Omega|},$$ where $E_k \subset \Omega$ is the set $$E_k := \{(D_1, D_2, D_3):\text{for each } 1 \leq i \leq 3, D_i \in \Bbb{N}, 1 \leq D_i \leq 6, \text{ and } D_1 + D_2 + D_3 = k \}.$$
Because $$(D_1, D_2, D_3) \in E_k$$ is equivalent to $$(7 - D_1, 7 - D_2, 7 - D_3) \in E_{21 - k},$$ we have $|E_k| = |E_{21-k}|$, so it suffices to evaluate $$E_3, E_4, ..., E_9, E_{10},$$ as we will then know the values of $$E_{18} = E_3, E_{17} = E_4, ..., E_{12} = E_9, E_{11} = E_{10}.$$ An application of stars-and-bars then gives us $$E_k = \binom{k-1}{2} - 3\binom{k-7}{2} = \frac{-2k^2+42k-166}{2} = -k^2+21k-83 = k(21-k) - 83$$ for $3 \leq k \leq 10$.
(The second term comes from excluding partitions of $k$ that have at least one part $>6$. We can partition $k-6$ into a sum of three positive integers in $\binom{k-7}{2}$ ways, and then we have $3$ choices of which positive integer to add $6$ back to to get a partition of $k$ where one part is strictly greater than $6$. This gives a total of $3\binom{k-7}{2}$ such partitions of $k$.)
This matches the original numerator when $k = 9, 10$. Since the quadratic $f(x) = x(21-x) - 83$ satisfies $f(x) = f(21-x)$, we know that this formula works for $k = 11, 12, 13, 14$ as well. Done.
