The 25 students in cs 70 have a strange way of picking a study group. each person is given a distinct number 1 through 25 based on random chance, and participants may form a group as cs 70, fall 2019, hw 7 long of the sum of their numbers is at most 162. (the tas are afraid of the number 162 for some how many different study groups are possible? that is, how many different subsets of students are there, which are allowable under the ranking rule? (hint: 162 = [( i)- 11/2) (b) uc berkeley is forming a new, n-person robotics team. there are 2n interested students n mechanical engineers and n programmers. (assume that no student is both a mechanical engineer and a programmer.) find the number of distinct n-person teams. (c) suppose that all robotics teams must also name a team captain who is a mechanical engineer. find the number of ways to pick an n-person team with a mechanical engineer as the captain.
4. picking teams (b) since each person on the team is unique we can simply ignore the fact that there are 2 professions. the amount of permutation of 2n students is 2n! , but since we are only picking n of them we will not need to pick anymore students after having n students on the team. hence the the amount of permutation of 2n students but only picking n student is (2n) (2n-1) (2n-2) (2n- (n) which is equal to 2 = 20). since the team is unordered, so a team consisting of student a, b, c is the same team consisting b, a, c we will need to divide the total amount of permutation by the number of orders. thus the total combination of 2n students to make a team of n student is (c) if a mechanical engineer is taken out of the pool of mechanical engineers, there will be n-1 of them left thus the total amount of students left to arrange is 2n - 1 students. applying the same process as part (b), we have 2n-1-11)1 = (n-1)! diving 1): = 2n-1). diving the total amount of permutation by the number of orders, we have 21 (12- ways.