Who ever gets all four of these right first wins! all problems are probability.

1. I run a book club with people, not including myself. Every day, for days, I invite members in the club to review a book. What is the smallest positive integer so that I can avoid ever having the exact same group of members over all days?

2. In how many ways can balls be placed in boxes if the balls are indistinguishable, and the boxes are distinguishable?

3. In how many ways can balls be placed in boxes if the balls are distinguishable, and the boxes are indistinguishable?

4. Six children are each offered a single scoop of any of flavors of ice cream from the Combinatorial Creamery. In how many ways can each child choose a flavor for their scoop of ice cream so that some flavor of ice cream is selected by exactly three children?

Sorry about my terrable spelling.

Geust Jul 22, 2024

#1**0 **

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.

bader Jul 23, 2024

#2**0 **

Problem 2:

There are 3 boxes, so there are 3 ways to place the first ball. There are 2 boxes left, so there are 2 ways to place the second ball.

There is 1 box left, so there is 1 way to place the third ball. There is 1 box left, so there is 1 way to place the fourth ball.

Therefore, there are 3 * 2 * 1 * 1 = 6 ways to place the 4 balls in 3 boxes.

magenta Jul 23, 2024

#3**0 **

Problem 3:

To solve this problem, we need to consider how distinguishable balls can be assigned to indistinguishable boxes. The key here is to count the distinct distributions of balls among the boxes without regard for which box is which.

Let's denote the balls as \(A\), \(B\), \(C\), and \(D\). Since the boxes are indistinguishable, we can only represent the counts of balls in each box. We can categorize the possible distributions based on how many boxes are used:

1. **All 4 balls in 1 box:**

- There is only one way to do this: \((4)\).

2. **3 balls in one box and 1 ball in another:**

- We can choose 1 ball to be separate. The number of ways to choose 1 ball out of 4 is \( \binom{4}{1} = 4 \).

- This distribution corresponds to \((3, 1)\).

3. **2 balls in one box and 2 balls in another:**

- We need to choose 2 balls to be together. The number of ways to choose 2 balls out of 4 is \( \binom{4}{2} = 6 \).

- This distribution corresponds to \((2, 2)\).

4. **2 balls in one box, 1 ball in a second box, and 1 ball in a third box:**

- First, we choose 2 balls to go in the first box, and the remaining two will each go in their own box. The number of ways to choose which 2 out of 4 balls go together is \( \binom{4}{2} = 6 \).

- This distribution corresponds to \((2, 1, 1)\).

Now, we can summarize the distributions:

- One way to have all balls in one box: \( (4) \)

- Four ways to have one box with three balls and another with one ball: \( (3, 1) \)

- Six ways to have two boxes with two balls in each: \( (2, 2) \)

- Six ways to have two boxes with one ball each and one box with two balls: \( (2, 1, 1) \)

Now we can tally these distinct distributions:

- 1 way for \( (4) \)

- 4 ways for \( (3, 1) \)

- 6 ways for \( (2, 2) \)

- 6 ways for \( (2, 1, 1) \)

To find the total number of ways to place the balls in the boxes, we add these up:

\[

1 + 4 + 6 + 6 = 17

\]

Therefore, the total number of ways to place 4 distinguishable balls into 3 indistinguishable boxes is \(\boxed{17}\).

magenta Jul 23, 2024