A person has 6 friends. Each evening, for 5 days, he or she invites 3 of them so the same group is never invited twice. How many ways are there to do this?
My approach is : Number of ways of selecting 3 friends out of 6 is 20 (6 choose 3). Now out of these 20 groups, for the first night I can choose any one group in 20 ways. For the second night I can choose a group in 19 ways and so on. So the answer should be 20+19+18+17+16 = 90. Although I am not sure about this.
Is this the correct solution or am I doing something wrong here?
P.S- this is question from mathematical circle and solution is not available for this.
The answer is $20\cdot 19\cdot 18\cdot 17\cdot 16$. To understand why just think about this, for EACH group you can choose the first day you have 19 options the day after (and so on), so you have $20\cdot 19$ ways of chosing the groups of the first two days.