$4$ judges must distribute grades from $1$ to $6$ for each participant. In how many ways one can score $22$ points?

Question

$4$ judges must distribute grades from $1$ to $6$ for each participant. To be a finalist, one must score $22$ points. Find the number of ways to attribute grades such that a participant it a finalist.

I wrote it as:

$$(x+\dots+x^6)^4=x^4(1+\dots+x^5)^4=x^4\left[\frac{1-x^6}{1-x}\right]^4=x^4 (1-x^6)^4\left[\frac{1}{1-x}\right]^4$$

Now I guess I just need to find the coefficient of $x^{18}$ in the expansion of $(1-x^6)^4\left[\frac{1}{1-x}\right]^4$.

Then I have:

$$(1-x^6)^4=\sum_{i=0}^4(-1)^i{4\choose i}x^{6i}\tag{A}$$

And:

$$\left[\frac{1}{1-x}\right]^4=\sum_{i=0}^{\infty}{4+i-1 \choose i}x^i\tag{B}$$

Looking at these formulas, I know that I just need to use $(A)$ and the answer should be:

$$A_0B_{18}+A_6B_{12}+A_{12}B_6+A_{18}B_{0}={4\choose 0}{21\choose 18}-{4\choose 1}{15\choose 12}+{4\choose 2}{9\choose6}-{4 \choose 3}{3\choose 0}=10$$

But the answer is wrong. I might have made some silly mistake that I couldn't spot.

Answer 1

I am pretty sure your calculation is right. Maybe the question is: In how many ways can a participant be a finalist (i.e. score at least 22 points). Then you need to add the coefficients of the $x^{19}$ and $x^{20}$ term as well, which are $4$ and $1$ respectively. I'm guessing the answer should be $15$?

Answer 2

Maybe easier method:

you are looking for numbers $a,b,c,d\in\{1,2,3,4,5,6\}$ such that $a+b+c+d=22$.

if $a,b,c,d\le 5$, then $a+b+c+d\le 20<22$.

so at least one of them must be $6$ and it's easy to see, that there are only two nonincreasing solutions: $6+6+5+5=22$ and $6+6+6+4=22$.

Now count all solutions: there are $\binom{4}{2}$ solutions of the first kind and $\binom{4}{1}$ solutions of the second kind, so $6+4=10$ in common.

but I'm also pretty sure that one becomes a finalist with score $\ge 22$ not only at $22$, then you have additional $\binom{4}{1}$ solutions $6+6+6+5$ and one $6+6+6+6$, so $15$ in common.

Answer 3

Suppose the marks allocated by the 4 judges are: a, b, c, d

$a + b + c + d = 22$

Each variable has a max limit of 6. Suppose a = a' less than 6, and so on and so forth with other variables. Then we can re-write the equation as:

$(6 - a') + (6 - b') + (6 - c') + (6 - d') = 22$

$a' + b' + c' + d' = 2$

Total solutions for above equation = $C(5,3) = 10$

If we care about achieving marks greater than 22, then applying same logic, we know that total ways for obtaining a score of 23 and 24 are C(4,3) and C(3,3) respectively.

Total combinations = $10 + 4 + 1 = 15.$