A math problem---easy?
Ok, so this is a little riddle I have been trying to solve, and as great as I am with abstract math, I have a hard time with more concrete things like this problem.
How many ways are there to arrange 1 million beans into 3 different colored bowls?
Note: The bowls are different, so one bean in the red bowl is different from one bean in the green bowl.
I understand to make smaller parameters (instead of 1 million) and that there is bound to be some algebraic formula to figure it out, but it is really bugging me that I am struggling with this!
Anyone have any ideas?? thanks!
Update:There can be 0 beans in one or 2 bowls, but not all 3
Update 3:I think I have a formula.
If n=number of beans, then ((n+1)(n+2))/2 is the number of ways. So that means with 1 million beans, there are 500,001,500,000 ways.
Yall's answers really helped me come to a conclusion. Thanks!
Comments
If each bean is unique, then there are 3^1,000,000 ways to put them into the bowls.
I suspect you are meant treat the problem as if each bean is not unique (so if you had 200,000 beans in bowl A, 300,000 beans in bowl B, and 500,000 beans in bowl C, and then you traded a bean in bowl A for a bean in bowl C, it would not count as a different solution because there are still the same number of beans in each bowl.)
Note: I am assuming there must be at least 1 bean in each bowl, because if there were zero beans in one bowl, the beans would really only be sorted into 2 bowls, not 3. If it is considered an acceptable solution to have 0 beans in a bowl, you'll have to modify the solution.
Let a, b, and c be the number of beans in bowls A, B, and C, respectively.
Start with bowl A...there can be from 1 to 999,998 beans in the bowl, leaving at least 1 bean in each of the other two bowls. So there are 999,998 possibilities for a.
Skip to bowl C for a minute: there must be 1,000,000 - a - b beans. So there is only one possibility for c for each combination of a and b. So we can ignore bowl C, and just focus on the possible combinations of a and b.
Bowl B: there can be up to from 1 to [999,999 - a] beans, leaving at least 1 bean for bowl C.
If a = 999,998 there is only one possibility for b: it must be 1.
If a = 999,997, there are two possibilities for b: it can be 1 or 2.
If a = 999,996, there are three possibilities for b: it can be 1, 2, or 3.
So the total number of possibilities so far is 1 + 2 + 3. If we went all the way down the list of possibilities for A, there would be a total of 1 + 2 + 3 + 4...all the way up to 999,998 (the total possibilities for A).
The formula for the sum of the first n integers is (n)(n+1)/2. So the answer is (999,998)(999,999)/2 = 499,998,500,001.
Let's think about this.
Suppose you put zero beans in the red bowl. Then there are 1,000,000 ways to divide the beans between the other two bowls.
Suppose you put one bean in the red bowl. Then there are 999,999 ways to divide the beans between the other two bowls.
Suppose you put two beans in the red bowl. Then there are 999,998 ways to divide the beans between the other two bowls.
If you think about this a little, you'll find that the total number of combinations is 1,000,000 + 999,999 + 999,998 + ... + 1
Now you use the trick for adding sequential numbers:
Two times this number is (1,000,000 + 1) + (999,999 + 2) + (999,998 + 3) + ... + (1 + 1,000,000).
Hence, two times the number of combinations is 1,000,000 * 1,000,001.
So the total number of combinations is 500,000 * 1,000,001
That's 500,000,000,000 + 500,000
or 500,000,500,000.
Hope that helps!