Class of 24 pupils and 4 chosen for committee members A class of 24 pupils consists of 11 girls and 13 boys. To form the class committee, four of the pupils are chosen at random as "Chairperson", "Vice-Chairperson", "Treasurer", and "Secretary". Find the number of ways the committee can be formed if
(i) the committee consists of at least one girl and at least one boy,
(ii) the "Treasurer" and "Secretary" are both girls,
(iii) The teacher requires a group of four students to represent the class in a student survey. Find the number of ways this group of students can be selected if there must be at least 1 girl and at most 2 boys.
My answers:
(i) GBBG, GBBB, GBGG
(11C2 x 13C2) + (11C1 x 13C3) + (11C3 x 13C1) = 9581
(ii) GGGG, BBGG, BGGG
11C4 + (11C2 x 13C2) + (11C3 x 13C1) = 6765
(iii) GGGG, BBGG, BGGG
11C4 + (11C2 x 13C2) + (11C3 x 13C1) = 6765
I have checked the correct answers, it shows that (i) 229944   (ii)  50820   (iii) 6765
I don't see my (i) is wrong, is it the correct answer for (i) of 229944 as incorrect?
For (ii) why it is using 11P2 x 22P2 = 50820 for the answer? Why this is a permutation question?
 A: i) you are close. In each case, once you have selected the boys and girls for the committee, you need to permute their assigned roles within the committee. Because individual boys and girls are still people, they are all distinct. So, you should multiply your answer by $4!$
ii) here, calculating the permutations of individuals to jobs is a bit tricky.
Case 1: $gggg$: you can freely permute the four girls.
Case 2: $bggg$: step 1: choose where the boy will go (chair or vice), then permute the three girls.
Case 3: $bbgg$: permute the boys among chair and vice, then permute the girls among treasurer and secretary.
iii) you got it correct. There are no "positions" in the group for the student survey, so no need to permute the students chosen.
A: As soon as you name the different positions in the committee, it's a permutation question, not a combination question. If Sally, Betty, Joe, and Mason are the four people on the committee, having Sally as Chair, Betty as Vicechair, Joe as Treasurer and Mason as Secretary is a different outcome from some other assignment of jobs to those same individuals. That's why the correct answer for part (i) is equal to your answer, times $4!$.
The only part of this question that calls for combinations, not permutations, is part (iii), because it's just a group of $4$ students being chosen, as opposed to four distinct positions being filled.
A: It wasn't clear at first that these are three different questions, rather than three conditions for the answer. Your teacher should write questions more clearly.
(i) There are $24P4$ ways to choose $4$ people from a set of $24$ where order matters (combinations are for when there aren't distinguished positions). There are $11P4$ ways to choose an all-girl committee, and $13P4$ ways to choose an all-boy committee. So there are $24P4-(11P4+13P4)$ ways to choose a committee that is neither all-girl nor all-boy.
(ii) There are $11P2$ ways to choose the Treasurer and Secretary. There are then $22$ people left to choose for $2$ positions, so $22P2$ ways to choose the remaining positions, giving $(11P2)*(22P2)$ in total.
