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Four fair six-sided dice are rolled. The probability that the sum of the results being $22$ is $\dfrac{X}{1296}$. The value of $X$ is _______

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There are only two possible sets whose elements sum to $22$ : $\{6,6,6,4\}, \{6,6,5,5\}$

Number of permutations for $1^{st}$ set $: \dfrac{4!}{3!} = 4$

Number of permutations for $2{nd}$ set $: \dfrac{4!}{(2!*2!)} = 6$

So total number of ways to sum $22 = 10$

So$X =10.$
by Boss (11.3k points)
edited
Using Generating Functions

$$x1+x2+x3+x4=22\ where\ 1\leq x_{i}\leq 6$$

Now we need to find Cofficent of $x^{22}$
$$=[X^{22}]\left \{ (x^{1}+x^{2}+x^{3}+x^{4}+x^{5}+x^{6})^{4} \right \} \\ =[X^{22}][x^{4}]\left \{ (1+x^{1}+x^{2}+x^{3}+x^{4}+x^{5})^{4} \right \} \\ =[X^{18}]\left \{ (1+x^{1}+x^{2}+x^{3}+x^{4}+x^{5})^{4} \right \} \\$$
$$=[X^{18}]\left \{ (1+x^{1}+x^{2}+x^{3}+x^{4}+x^{5})^{4} \right \} \\ =[X^{18}]\left ( \frac{1-x^{6}}{1-x} \right )^{4}\ using\ GP\ Formula \\ =[X^{18}]\left ( \frac{1}{1-x} \right )^{4} \left ( 1-x^{6} \right )^{4} \\ =[X^{18}]\underbrace{\left [ \sum_{r=0}^{\bowtie } \binom{4+r-1}{r}x^{r}\right ]} \underbrace{\left [ \sum_{r=0}^{4}\binom{4}{r} (-x^{6})^{r} \right ]} \\$$
IN First and Second Part we can put the values as
$$r=18 \ r=0 \\ r=12 \ r=1 \\ r=6 \ r=2 \\ r=0 \ r=3$$
$$=\binom{21}{18}x^{18}-\binom{15}{12}\binom{4}{1}x^{18}+\binom{9}{6}\binom{4}{2}x^{18}-\binom{3}{0}\binom{4}{3}x^{18} \\ =(1330-1820+504-4) x^{18} \\ =10x^{18}$$
So, there are $10$ ways to get sum as $22$.

I know this method seems tedious and long, but it is mechanical and universal if used carefully.
by Active (4.8k points)
edited
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@bhuv

Sir Can u please tell the online source or book for learning Generating Functions?
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@bhuv very good analysis. But small correction is required in your explanation (check for typo)

+1
Thnx. For pointing out. Corrected now.
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good work @bhuv
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$\dfrac{1}{(1-x)^4}:\dfrac{1}{(1-x)^n}=\sum_{k=0}^{\infty }\ ^{n-1+k}C_{k}\ x^k$

$(1-x^6)^4:(1-x^r)^n=\sum_{k=0}^{n}\ ^{n}C_{k}\ (-x^r)^k$

In general, Probability (of an event ) = No of favorable outcomes to the event / Total number of possible outcomes in the random experiment.

Here, 4 six-faces dices are tossed, for one dice there can be 6 equally likely and mutually exclusive outcomes. T

aking 4 together, there can be total number of 6*6*6*6 = 1296 possible outcomes.

Now, No of favorable cases to the event : here event is getting sum as 22. So, there can be only 2 cases possible.

Case 1: Three 6's and one 4, for example: 6,6,6,4 ( sum is 22) Hence, No of ways we can obtain this = 4!/3! = 4 ways ( 3! is for removing those cases where all three 6 are swapping among themselves)

Case 2: Two 6's and two 5's,for example: 6,6,5,5 ( sum is 22) Hence, No of ways we can obtain this = 4! /( 2! * 2!) = 6 ways ( 2! is for removing those cases where both 6 are swapping between themselves, similarly for both 5 also)

Hence total no of favorable cases = 4 + 6 = 10. Hence probability = 10/1296. Therefore option D.
by Loyal (9.7k points)
–1 vote
Why we r not thinking this way -

(6,6,6,4) OR (6,6,5,5)

So probability will be:

(1/6 * 1/6 * 1/6 * 1/6) + (1/6 * 1/6 * 1/6 * 1/6) = 2/1296

So, value of X is 2
by Junior (883 points)
+6
6 6 6 4 can occur in 4C3 = 4 ways. (6,6,6,4), (6,6,4,6), (6,4,6,6), and (4,6,6,6)

Similarly, 6,6,5,5 can occur in 4C2 = 6 ways

So, 6+4 = 10.