One way is to find the coefficient of in the expansion of , which can be done in WolframAlpha. I would not say that this restatement makes the counting easier; it just reduces the problem to a well-known question for which automatic tools exist.
One way is to find the coefficient of in the expansion of , which can be done in WolframAlpha. I would not say that this restatement makes the counting easier; it just reduces the problem to a well-known question for which automatic tools exist.
OK, its nice and all, I have tried to solve it in a different way that seems so correct but doesn't work and I'm confused. Look what I did:
because xi >= 1 we can change variable to xi = yi + 1 then we have y1 + 1 + y2 + 1 + y3 + 1 + y4 + 1 = 10 => y1+y2+y3+y4= 6. now we want to find the number of solutions to y1+y2+y3+y4= 6 where for every i, 0<=yi<=4.
for getting the number of solutions to the latter equation where yi >=0, we can use the formula for dividing 6 identical objects into 4 different boxes it's |U|= nCr(4-1+6,4-1)=84, the size of our universe. now we can use the inclusion & exclusion in order to find number of solutions where y1 >= 5 or y2 >= 5 or y3 >= 5 or y4 >= 5, there are only 16 such solutions. so the final answer is 84-16=68. but the answer is incorrect. where is my mistake? I cant figure it out, just looking for it for hours.
It should be, 0 <= y_i <= 3.because xi >= 1 we can change variable to xi = yi + 1 then we have y1 + 1 + y2 + 1 + y3 + 1 + y4 + 1 = 10 => y1+y2+y3+y4= 6. now we want to find the number of solutions to y1+y2+y3+y4= 6 where for every i, 0<=yi<=4.
Your method is otherwise correct. I was just looking into it at this thread. Let s be the sum of all 's, namely, 6, and let m be the maximum of 's, namely, 3. What makes it easy to subtract the number of variants where some is that . In this case, when , , so the restriction that does not have to be taken into account and the number of solutions of can be calculated by the general formula.
By the Erdos principle, there is always a simple explanation for a simple answer
Ginsburg was quite close in an earlier post.
Number of solutions to
with
is the number of solutions to (substitue )
with
Number of solutions to
with
is
We need to throw away solutions where one of .
For the case of , let , the number of solutions to
is
Since exactly one of could be larger than 3 in any solution.
We have .
Subtracting invalid solutions, the number of possible solutions is:
Consistent with awkward's solution.
CB, I find 44 solutions, listed below. Do you differ with that list?
Code:1 1 1 4 4 2 1 2 3 4 3 1 2 4 3 4 1 3 2 4 5 1 3 3 3 6 1 3 4 2 7 1 4 1 4 8 1 4 2 3 9 1 4 3 2 10 1 4 4 1 11 2 1 3 4 12 2 1 4 3 13 2 2 2 4 14 2 2 3 3 15 2 2 4 2 16 2 3 1 4 17 2 3 2 3 18 2 3 3 2 19 2 3 4 1 20 2 4 1 3 21 2 4 2 2 22 2 4 3 1 23 3 1 2 4 24 3 1 3 3 25 3 1 4 2 26 3 2 1 4 27 3 2 2 3 28 3 2 3 2 29 3 2 4 1 30 3 3 1 3 31 3 3 2 2 32 3 3 3 1 33 3 4 1 2 34 3 4 2 1 35 4 1 1 4 36 4 1 2 3 37 4 1 3 2 38 4 1 4 1 39 4 2 1 3 40 4 2 2 2 41 4 2 3 1 42 4 3 1 2 43 4 3 2 1 44 4 4 1 1