Here's the solution to problem 1:
To avoid getting the same group of 5, we need to ensure there are enough total members such that any group of 5 can be chosen without repeating a combination.
Here's the key idea: We care about the number of distinct groups of 5 members we can form, not just the total number of possible selections.
We know the difference between combinations and permutations. Combinations focus on the group itself, regardless of order (e.g., John, Mary, Sarah is the same group as Sarah, John, Mary). Permutations consider order (e.g., John reviewing first is different from Mary reviewing first).
In this case, since order doesn't matter, we're interested in the number of combinations of 5 members we can choose from a group of n members. This is denoted by n choose 5, written as nC5 or (5n).
For 400 days, we want enough combinations to ensure no repetition. If we have nC5≥400, then we're guaranteed to have at least 400 distinct groups of 5 members.
Now, calculating nC5 directly can be cumbersome for large numbers. Thankfully, there's a property that helps: $_n\text{C}_k = nCn−k (combinations are symmetrical). This allows us to consider the smaller value of k, which in our case is k = n - 5.
Therefore, we want nCn−5≥400.
Finding the smallest n that satisfies this inequality can be done through trial and error or using a calculator with a combinations function. Experimenting with small values of n, we find that:
8C3=56<400 (not enough combinations)
9C4=126≥400 (enough combinations)
Therefore, the smallest positive integer n that allows for 400 days of non-repeating book club groups of 5 is n = 9.
