Easy math puzzles

Quote: Wizard

Here is something a bit easier.

How many ways are there to put six distinct balls into three identical boxes?

Is there a general formula for n distinct balls into three identical boxes?
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There are only seven partitions of six using 1-3 digits:

6
15
24
33
114
123
222

It's easy to calculate the number of combinations for any of them. For instance, there are combin(6,1) * combin(5,2) * combin(3,3) = 60 combinations of 123. The combinations of all seven partitions sum to the answer of 122 combinations for boxes containing 0-6 balls.

If all boxes must contain at least one ball, then we subtract the 32 combinations of 6, 15, 24 and 33 to get the answer of 90 combinations

Incidentally, regarding the “random board cuts” problem: You guys analyzed it for days, eventually concluding there was no direct solution for n cuts, then I post the direct solution using nothing but basic combinatorics/calculus, and apparently no one cares?

Quote: Ace2

The combinations of all seven partitions sum to the answer of 122 combinations for boxes containing 0-6 balls.

I agree!

Sorry for the tardy response.

Quote:

Incidentally, regarding the “random board cuts” problem: You guys analyzed it for days, eventually concluding there was no direct solution for n cuts, then I post the direct solution using nothing but basic combinatorics/calculus, and apparently no one cares?
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Sorry but I was out of town for almost a month and didn't see it. I'll have a look.

Quote: Ace2

Let's say you have 100 baseballs lined up on the ground and you randomly divide them into 5 groups (a group can contain 0-100 balls). There are combin(101,4) ways to insert 4 dividers among 101 positions to create the 5 groups. If you want to ensure all 5 groups contain at least 3 balls, then you remove 5*3 balls, then randomly insert 4 dividers into the remaining 85 balls ((combin(86,4) ways), then add 3 balls back to each group. Therefore, the probability that each of the 5 random groups contains at least 3 balls is 86*85*84*83/(101*100*99*98). As the total number of balls (n) goes to infinity, the probability that each group contains at least 0.03n balls is obviously 0.85^4. Therefore, as n goes to infinity, the probability that all 5 groups contain least xn balls is (1 - 5x)^4

So if we make 4 random cuts into a 1 foot long board, we know that the probability of all 5 pieces being at least x long is (1 - 5x)^4. Now we apply the very useful property that the expected value of an event to happen is the cumulative probability of it never happening. The "event" is finding a piece shorter than x. For this problem, this means: add the probabilities that all pieces are longer than x for all x from 0 to 0.20 (obviously all 5 pieces can't be longer than 0.20). We get this sum by integrating (1 - 5x)^4 dx from 0 to 0.20, which gives us the answer of 1/25 feet =expected length of shortest piece

Putting it all together, the expected length of the shortest piece when making (s - 1) cuts into a 1 foot board to get (s) pieces is the integral from 0 to 1/s of (1 - sx)^(s-1) dx, which evaluates to 1/s^2. So, for example, if we are cutting the board into s=8 pieces, the expected length of the shortest piece is 1/8^2

Note: the probability of all segments having a minimum size can be explained without the discrete baseball example as follows. But IMO, the discrete example is more intuitive. Starting with a board 1 foot long, you want to make 4 random cuts into 5 pieces with a minimum length of 0.03. You accomplish this by drawing a line at length 5 * 0.03 =0.15, then make the 4 random cuts irrespective of the line. If they all fall on the 0.85 side then subdivide the 0.15 side into 5 pieces of 0.03 and fasten one of them to each of the 5 pieces cut on the 0.85 side. So there's a (1-0.15)^4 chance they all fall on the 0.85 side and no pieces are <0.03
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This is really good. Thank you. I think I'll make an Ask the Wizard question out of it. Sorry it took so long to reply. I was out of town.