FOR A
a) Assume for contradiction that the class has less than 5 male AND less than 5 female students.
Then the number of students in the class is:
C = M + F </= 4 + 4 = 8
But there are 9 in the class.
Hence the above assumption is false so statement A is true,
FOR B
Assume for contradiction that the class has less than 3 male AND less than 7 female students.
Then the number of students in the class is:
C = M + F </= 2 + 6 = 8
But there are 9 in the class.
Hence the above assumption is false so statement B is true,
The pigeonhole principle is basically a counting argument that says if you have n items to put into m pigeonholes with n > m, at least on pigeonhole must contains more than one item.
In this particular case, there are 9 students (items) and 2 pigeonholes (genders), so when you have assigned gender to 8 students, you either already have a group (pigeonhole) with 5 students of the same gender, or you have two groups, each with four students. Therefore, regardless of the gender of the 9th student, one of the groups must end up with 5.
Similarly, after you've assigned gender to 6 students, either there were already 3 males, or there were 0, 1 or 2 males. You still have 3 more students, so if there were 0 males, there must have been 6 females, and either the 3 remains students are all males (in which case, there will be 3 males) or one of the is a female (in which case, there will be at least 7 females). The argument for 1 or 2 males out of the 6 is similar.
https://math.stackexchange.com/questions/443808/pigeonhole-principle-show-that-a-class-of-nine-has-at-least-five-male-or-five-f
