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How many solution does $x+y+z=11$ have where $x, y, z$ are non-negative integers. In light of the restrictions, its clear that $x,y,z \in \{0,1,2,..11\}$. So, at face value I would assign a value for $x$ and determine the different combinations that $y$ and $z$ can hold. For example,

For $x=0$, we have $y+z=11$. With writing them out I found that there are $12$ different assigned combinations for $y$ and $z$ that satisfy the equation. For $x=1$, I got $11$. Consequently, the pattern becomes clear whereby each one takes a value less by one. Hence, the number of solutions is $1+2+3+4+5+6+7..+12=78$. I was wondering if there is an easier method perhaps with combinations equation $C(a,b)$..?

John
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  • We have had too many questions of this type. – evil999man May 03 '14 at 12:13
  • See my answer here:http://math.stackexchange.com/questions/689975/distributing-n-different-things-among-r-persons/690036#690036 – evil999man May 03 '14 at 12:15
  • Guys, stop upvoting duplicate homework questions, seriously! – Alec Teal May 03 '14 at 12:18
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    @Alec Lighten up. The OP shows plenty of effort, and arrives at the correct answer in doing so. It deserves an upvote. John likely didn't know that this is a classic sort of problem, and probably hasn't encountered it before. Nor that many similar questions can be answered by the same method. And if you believe it is a duplicate and should be slammed shut because of it, then why'd you answer it? Besides, stop playing the homework police on questions showing commendable levels of effort! – amWhy May 03 '14 at 12:24
  • @amWhy Not thinking something deserves an upvote isn't the same as saying it deserves to be closed. When I upvote something, what I personally mean is "this is the kind of content I come to the site for". My preferred policy would for routine problems to be dealt with quickly and without fanfare (positive or negative), and to save upvotes (and therefore time on the front page) for the more original, thought-provoking questions. – Jack M May 03 '14 at 12:28
  • @JackM You are fully entitled to vote as you described. But that shouldn't be taken to say that "all folks should vote like this" or that "my policy should be everyone's policy." – amWhy May 03 '14 at 12:30
  • @AlecTeal I have to ask you a question? Based on what premise, did you deduce that this is a homework question? Unfortunately, everyone(well let me say most of the people) on this website jumps to false conclusions and make hasty generalizations, and I wonder how do they get to it. Please enlighten me. If things aren't clear enough, I must say that this isn't a homework problem, but I doubt it could change your mind because clearly its made up. – John May 03 '14 at 13:07
  • @Awesome I just realized that there are many similar questions, although I made a previous search. In any case, though I have flagged my question as a duplicate. – John May 03 '14 at 13:30
  • @John because homework questions (or stuff from books) are on the same lines and it has been discussed to death. I remember my combinatorics assignments..... :P – Alec Teal May 03 '14 at 17:01
  • @AlecTeal Let me clarify that a question does not necessarily have to be a homework question. As such, I recommend not making any false assumptions and inductive fallacies; it doesn't hurt to ask the OP. Furthermore, if it was a homework question, now would I be here if I had the answer. Ultimately, my goal in asking the question is to understand and I have done so. It would be most unproductive to label future questions "homework" especially here where it unfortunately takes a pejorative tone. – John May 03 '14 at 20:12
  • @AlecTeal Homework or not, a question is a question. The final goal is to understand and you or anyone for that matter shouldn't stop the OP from learning. – John May 03 '14 at 20:12
  • @AlecTeal On that note, I appreciate the answer you gave below, it says that you want to help, which is a good thing. – John May 03 '14 at 20:19

3 Answers3

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This is a version of the classic stars-and-bars problem in combinatorics.

For any pair of natural numbers $n$ and $k$, the number of distinct $n$-tuples of non-negative integers whose sum is $k$ is given by the binomial coefficient $$\binom{n + k - 1}{k}$$

Here, $n = 3$, and $k = 11$, giving you $$\binom{3 + 11 - 1}{11} = \binom{13}{11} = \dfrac{13\cdot 12}{2} = 6\cdot 13 = 78$$

amWhy
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  • The link takes you to Wikipedia's Stars and Bars entry, where you'll find a nice explanation of why this works, using an example that's fleshed out nicely. – amWhy May 03 '14 at 12:17
  • Using the formula makes things easy, but I don't see what the connection is. Even with using the stars and bars method, it seems rather inappropriate, due to the dependency of the variables. In other terms, if x takes a value it creates a dependency on the other variables. I would appreciate it if you could make the connection clearer. – John May 03 '14 at 13:41
  • Did you read the Wikipedia entry? It elaborates on the "why's" of this formula. – amWhy May 03 '14 at 13:45
  • I read it and I understood everything but when I went to this question, http://math.stackexchange.com/questions/322369/how-many-non-negative-integral-solutions-does-the-equation-x-1-x-2-x-3-10, with the additional restrictions I started to doubt everything. None of the answers that used combinations were clear. – John May 03 '14 at 13:49
  • First get comfortable without the restrictions, getting a firm grasp on the non-negative integers, and the number of positive integers. Then we can begin adding restrictions. – amWhy May 03 '14 at 13:53
  • Another confusing thing for me was the zero. I was considering it to be an element, i.e. to be represented by a star. But, after thinking about it I figured it out. I still have to figure out a greater than restriction like $x>2$ – John May 03 '14 at 14:27
  • Well, taking a simple example: $x_1 + x_2 = 11$, where the $x\gt 2$, we can simply replace each $x_i$ with $x_i = y_i + 2$ and solve $y_1 + 2 + y_2+ 2 = 11 \iff y_1+y_2 = 7$. Then we have the case in which we are seeking the number of positive integer solutions to $y_1 + y_2 = 7$. – amWhy May 03 '14 at 14:34
  • I deleted my comment, but let me re-add as I think its correct. I said we first initialize everything so if we have $x>=2$. We have to put two stars for x as such: **|-|- . Now the remaining stars are $11-2= 9$. And to put 9 indistinguishable objects in 3 distinct boxes we need $C(3+9-1,9)$ – John May 03 '14 at 14:39
  • Its correct, right? – John May 03 '14 at 14:42
  • You'd need to subtract from you total $11$, two stars for each $x_i$, giving you $5$ stars to work with. For example, for $x_1 + x_2+x_3 = 11$, putting $x_i = y_i + 2$ gives us $y_1 + 2 + y_2 + 2 + y_3 + 2 = 11 \iff y_1 + y_2 + y_3 = 5$. Then the number of integer solutions greater than or equal to $0 = \binom{5+3 - 1}{5}$ – amWhy May 03 '14 at 14:43
  • Are you sure because http://math.stackexchange.com/questions/322369/how-many-non-negative-integral-solutions-does-the-equation-x-1-x-2-x-3-10 here the suggested and thereafter confirmed answer is $C(3+8-1,8)$. Its 8 because instead of 11, its 10. So, $10-2=8$ – John May 03 '14 at 14:46
  • Are you still there? Is the solution to that problem wrong? – John May 03 '14 at 14:57
  • In that problem, it specifies only one variable $x$ being greater than or equal to two. My work shows what you get when the restriction is that all the variables must be integers greater than or equal to 2. So the principal is the same. For each variable needing to be $\geq 2$, we subtract 2 for each such variable. – amWhy May 03 '14 at 15:09
  • So, what I wrote is correct since I specified $x\ge 2$ and not $y$ or $z$ – John May 03 '14 at 15:13
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    Yes. I see you used $x, y, z$, so if only $x$ needs to be greater than or equal to 2, then the solution (with k=11 - 2 = 9) is $\binom{3 + 9 - 1}{9}$. – amWhy May 03 '14 at 15:17
  • You're welcome, John! – amWhy May 03 '14 at 16:09
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imagine 11 balls in a row and two blocks which you will place somewhere. you insert the blocks before, after or between the balls and then you assign values to $x,y,z$ in the following way: $x$ is the number of balls from the beginning of the row up to the first block, $y$ the number of balls between the two blocks and $z$ number of balls from the second block up until the end of the row. you will easily see that the number of ways in which you can place the blocks is equal to the number of different triplets $x,y,z$. Do you know how to compute the number of possible distributions of blocks?

mm-aops
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Okay let us write a solution to $a+b+c+d+e=10$ a different question, just incase it is homework.

Each solution will have the form:

||||-|---||||| <-> 4As 1B 0Cs, 0Ds, 5Es

How many different ways can we arrange 10 |s and 4 (4=5-1) -s? Each arrangement of these |s and -s is a valid solution.

$$\frac{(10+4)!}{4!10!}=\frac{14!}{10!4!}$$

Alec Teal
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