在卡里斯马初级学院（Charisma Junior College）的学生会中，共有 31 名成员，其中包括学生服务（SS）部门的 22 名成员和魅力影响力（CI）部门的 9 名成员。在学院的晨会上，共有 6 名成员值日，两人一组进行。 每天早晨的三项不同值日任务包括：每日启示（Daily Inspirations）、升旗（Flag Raising）以及领唱国歌和领读誓词（Recital of the national anthem and pledge）。除了“每日启示”部分必须由 CI 部门的成员执行外，所有值日任务都可以由任何学生会成员执行。求：

1. Find the number of ways to assign members to the Daily Inspirations duty.

The Daily Inspirations duty requires 2 members, both of whom must be from the CI Wing (which has 9 members).
Number of ways = $\binom{9}{2} = 36$

2. Find the number of ways to assign the remaining members to the other two distinct duties and calculate the total.

The remaining two duties (Flag Raising and Recital) are distinct and require 2 members each. There are no restrictions, so we choose from the remaining $31 - 2 = 29$ members.
Number of ways for Flag Raising = $\binom{29}{2} = 406$
Number of ways for Recital = $\binom{27}{2} = 351$
Total number of possible duty groupings = $36 \times 406 \times 351 = 5\,130\,216$

1. Calculate the total number of possible duty groupings (using the result from part i).

To find the total unrestricted groupings, we choose 2 members from the 9 CI members for Daily Inspirations (DI). Then, from the remaining 29 members (31 total - 2 chosen for DI), we choose 2 for Flag Raising (FR). Finally, from the remaining 27 members, we choose 2 for Recital (R).
$\text{Total groupings} = \binom{9}{2} \times \binom{29}{2} \times \binom{27}{2}$
$= 36 \times 406 \times 351 = 5,130,216$

2. Calculate the number of duty groupings where Jill is NOT on duty.

If Jill is entirely excluded from the morning duties, we remove her from the available pool. The CI Wing now has $9 - 1 = 8$ members, and the overall available pool (excluding Jill) is $31 - 1 = 30$ members.

We choose 2 for DI from the remaining 8 CI members:
$\binom{8}{2} = 28$
The remaining available pool is now $30 - 2 = 28$ members. We choose 2 for FR:
$\binom{28}{2} = 378$
The remaining pool is $28 - 2 = 26$ members. We choose 2 for Recital:
$\binom{26}{2} = 325$

$\text{Groupings without Jill} = 28 \times 378 \times 325 = 3,439,800$

3. Subtract the groupings without Jill from the total groupings to find the answer.

Using the complement method, the number of groupings where Jill is on duty is the total possible groupings minus the groupings where she is excluded:
$\text{Groupings with Jill} = 5,130,216 - 3,439,800 = 1,690,416$

1. Establish Jack's possible duties and calculate the ways to assign the Daily Inspirations (DI) duty.

Since Jack is in the SS Wing, he cannot be assigned to the Daily Inspirations (DI) duty (which is restricted to CI members). He must therefore be assigned to either the Flag Raising (FR) or the Recital (R) duty.

The DI duty must first be filled by 2 members chosen from the 9 in the CI Wing:
$\text{Ways to assign DI} = \binom{9}{2} = 36$

2. Assign Jack to a duty, choose his partner, and assign the remaining duty to find the total combinations.

Jack can be assigned to either the FR or R duty ($2$ choices).
Once his duty is chosen, he needs $1$ partner from the remaining $28$ members (since $2$ are in DI and Jack is $1$, leaving $31 - 2 - 1 = 28$): $\binom{28}{1} = 28$.

The remaining duty requires $2$ members from the $27$ members left: $\binom{27}{2} = 351$.

Using the multiplication principle:
$\text{Total groupings} = 36 \times 2 \times 28 \times 351 = 707,616$

1. Establish the values from the earlier parts to compare the number of groupings for a CI member versus an SS member.

Total possible duty groupings (from part i):
$N = \binom{9}{2}\binom{29}{2}\binom{27}{2} = 36 \times 406 \times 351 = 5,130,216$

Groupings where Jill, a CI member, is on duty (from part ii), found by subtracting groupings without her:
$N - \binom{8}{2}\binom{28}{2}\binom{26}{2} = 5,130,216 - (28 \times 378 \times 325) = 1,690,416$

Groupings where Jack, an SS member, is on duty (from part iii), found by subtracting groupings without him:
$N - \binom{9}{2}\binom{28}{2}\binom{26}{2} = 5,130,216 - (36 \times 378 \times 325) = 707,616$

2. Compare the number of groupings to justify the comment on fairness.

Since $1,690,416 > 707,616$, a CI member is involved in more than double the possible duty groupings compared to an SS member. If a grouping is picked uniformly at random, a CI member inherently has a higher probability of being chosen for duty. Thus, the system is not fair.