Question

Suppose that a non-negative integer solution to the equation w + x + y + z = 10 is chosen at random (each one being equally likely to be chosen). What is the probability that in this particular solution w is less than or equal to 2?

Could somebody help me determine how to find the total number of combinations? (Don't just tell me the answer, I need to know how) @Discrete Mathemat…

Answer 1

The general approach for these kinds of problems is to first find the universe of events (the solutions to the equation), then find how many satisfy the given criteria (how many have w <= 2). Then divide.

Answer 2

any more insight on finding the amount of events that satisfy the criteria? that's where i'm having trouble. i still cannot solve this problem

Answer 3

They must be non-negative, so they can be 0. But none can ever be more than 10. And the expression is symmetric in the 4 variables so you can treat them all equally.

Not all 4 can be 0, but 3, 2, 1, or 0 can be 0. I'd start by breaking the problem down into those 4 cases. The case where 3 are 0 is the easiest.

3 0's:
The other value is necessarily 10. There are 4 variable which can be 10, then the others are all 0. So there are 4 ways to assign the values. 4 total solutions.

2 0's:
Two variables are 0, so the other two must add to 10. Once you know one, you know the other. There are 10 ways (ignoring the case of 0, which was already counted in the previous case) to assign the value of one variable, and each way dictates the value of the other variable. However, we have to be careful because, for example (w, x) = (5, 5) is really the same case as (x, w) = (5, 5). There are 4 choose 2 (4C2) = 6 ways to pick the two nonzero variables. For each of those ways there are 10 ways to set the values for the variables. So 60 solutions. But 6 of those will be repeated (5, 5) solutions, so really 54 total solutions.

1 0:
There are 4C3 = 4 ways to pick the 0 variable. The other three are nonzero. I'd probably approach this one by letting one value be 1 through 8 (because the other two have to be at least 1), then treating each of those cases like the above case of 2 nonzero variables.

0 0's:
Same as above, but hairier.

I don't know of a neat formula to apply for these kinds of problems. It's really just brute force with a little logic to simplify the work.

Another way is to use the 4 cases (0, 1, 2, or 3 0's), then write down tuples that add to 10, with non-decreasing values in the tuples. For example, in the 0 0's case:
(1, 1, 1, 7)
(1, 1, 2, 6)
(1, 1, 3, 5)
(1, 1, 4, 4)
(1, 2, 2, 5)
(1, 2, 3, 4)
(1, 3, 3, 3)
(2, 2, 2, 4)
(2, 2, 3, 3)
Actually, that might be all the possible tuples. But they can't each be assigned to (w, x, y, z) in 4! = 24 ways. The first one can only be done in 4 ways. The next one in 12 ways. The fourth one in 6 ways. Then do the same exercise for the cases with some variables set to 0.

Once you get the cases enumerated, figure out how many of each example can have w <= 2. In the examples above, the fourth one can works for half of its permutations. But the first example works for 3 of 4 permutations.

Answer 4

I thought of it differently, although I'm not exactly sure if I have the correct answer..

The total number of combinations for w+x+y+z=10 is C(13,3) which is found from using the method of binary sequences. This looks at a string of 0's and 1's where the 1's act as separators for 0's which represent the value of w,x,y,z.

For instance, if w=1, x=2, y=3, z=4 the string would be: 0100100010000
Finding the possible ways this string can be made is the same as asking "how many 13 digit binary strings can be made with three 1's" which comes out to C(13,3).

To find how many instances w less than equal 2, I considered the three cases:
i) w=2 means x+y+z=8
ii) w=1 means x+y+z=9
iii) w=0 means x+y+z=10

The possible ways these three cases can occur is C(10,2) + C(11,2) + C(12,2)
So the probability of w less than equal 2 is (C(10,2) + C(11,2) + C(12,2))/C(13,3).

Answer 5

@bananaradiation That's brilliant. I'll probably work it out both ways to make sure, but I can't see why you wouldn't be right.

Answer 6

@bananaradiation - Thanks a lot. I didn't actually read your whole answer, but I managed to find the method for solving these kinds of problems in my book, and the correct way to approach this is with a binary string like I saw in your answer.