Question

5. Counting problems, leave answers as expressions, e.g., 10 npr 4, 8 nCr 3 or 35 [5 pts each part]

a) A club elects a steering committee of 5. Among these 5, a chair and secretary are chosen. How many different sets of committees (including officer selection) are possible if club has 15 members? Officers are not the same as the non-officer members, so this is hybrid permutation/combination problem.
b) At an event that drew 100 people where each attendee gets one raffle ticket, there are 6 raffle prizes worth $200, $100, $50, $25, $25, $25. How many different raffle ticket winner selections are possible? Be careful, the $25 prizes are equivalent, so this is hybrid permutation and combination problem.
c)If repeats are allowed but a code cannot begin with 0, how many six-digit PIN codes can be made?​

Answer 1

Explanation:

a) The steering committee of 5 can be chosen from 15 members in 15 nCr 5 ways. Once the committee is chosen, the chair can be selected in 5 ways and the secretary can be selected in 4 ways (since the chair cannot also be the secretary). Therefore, the total number of different sets of committees (including officer selection) is:

15 nCr 5 * 5 * 4 = 3,003,600

b) Each of the 6 raffle prizes can be awarded to any of the 100 people, so there are 100 choices for the first prize, 99 choices for the second prize, and so on, down to 95 choices for the sixth prize. However, since the three $25 prizes are equivalent, we need to divide by 3! to account for the ways in which the $25 prizes can be arranged. Therefore, the total number of different raffle ticket winner selections is:

(100 * 99 * 98 * 97 * 96 * 95) / (3!) = 903,450,240,000

c) Since repeats are allowed and the code cannot begin with 0, there are 9 choices for the first digit and 10 choices for each of the remaining digits. Therefore, the total number of six-digit PIN codes that can be made is:

9 * 10^5 = 9,000,000