Finding the values for satisfying the given equality $x+y+z=7$

Question

If $x,y,z$ are three given positive integers such that they are never less than $1$(but can be equal to one) where $x+y+z=7$ ..... then prove that all the possible solutions of $x,y,z$ are contained in the set $(1,2,4)$ when no two values are same (of $x,y,z$ that is for example $1,3,3$ is not a solution because two values are equal)....

I solved this question by trial and error and in the process wrote all the possible values(which is indeed a very bad solution)..but still I solved this question but ...i got this question in combinatorics/Binomial theorem so I thought there must be some solution(probably related to these)... better than writing all the possibilities ...which I have not been able to get...any help is appreciated...

Answer 1

If no two values are the same, we can order them in increasing order. We also know that they are all positive integers. So we can say, without loss of generality, that

$$1\le x<y<z$$

Now let's look for solutions in increasing lexigraphical order: i.e. $(x_1,y_1,z_1)$ comes before $(x_2,y_x,z_2)$ iff $x_1<x_2$ or ($x_1=x_2$ and $y_1<y_2$) or ($x_1=x_2$ and $y_1=y_2$ and $z_1<z_2$).

The smallest possible value of $x$ is $1$, and then the smallest possible for $y$ is $2$. This makes $z=4$, which is the given answer.

The next smallest is $x=1,y=3$, but this makes $z=3$, and the values are not distinct. Increasing $y$ makes $z$ smaller, so we cannot have $x=1,y=3$.

The next smallest is $x=2$. That makes the smallest $y$ to be $3$, and thus $z$ is at most $2$--smaller than $y$ so not allowed. This will be the case for any larger values for $x,y$.

Thus there is only one solution for $1\le x<y<z$ and $x+y+z=7$.

Answer 2

$$x=a+1,y=b+1,z=c+1$$ $$a+b+c=4$$ $$\text{Assuming }a\leq b\leq c$$ $$(a,b,c)=(0,0,4),(0,1,3),(0,2,2),(1,1,2)$$ $$\therefore \{a,b,c\}=\{0,1,3\}$$ $$\therefore \{x,y,z\}=\{1,2,4\}$$

Answer 3

The total number of non-negative solutions is ${(7-1) \choose(3-1)}=15$ and the solutions which are repeated are $(3,2,2),(1,3,3),(1,1,5)$ no other positive solutions are repeated. So the asolutions are $15-(3.\frac{3!}{2!})=6$ . Now going for other ones is easy rather than for 15. Hope its clear.thats just the help for trial and error method.

Answer 4

Rory Daulton has already given an excellent answer using algebra and logic.

A slightly different way to approach a problem like this is to visualize it as a problem of putting $7$ identical balls in to three boxes labeled $x$, $y$, and $z$.

The restriction that all three values must be positive is equivalent to saying we must put at least one ball in each box. So let's do that. Now we have four balls left to put in three boxes. (This is the same simplification that Kay K. performed.)

We cannot put all the remaining balls in one box, because that leaves the other two boxes with the same number of balls. So we must add at least one ball each to at least two boxes. Without loss of generality, then, suppose we put a ball in box $y$ and a ball in box $z$.

Now boxes $y$ and $z$ still have the same number of balls, so we will have to put at least one ball in one of them in order to make them different. Without loss of generality, suppose we put another ball in box $z$.

Now the boxes contain $1$, $2$, and $3$ balls respectively, and we have one remaining ball to place. It cannot go in either of the first two boxes (because that would make two boxes the same), hence it must go in the last box, which now contains $4$ balls.

So we end up inevitably with $1$, $2$, and $4$ balls in the boxes. That is, these are the only three distinct positive integers whose sum is $7$.