1. ## Concept Clarification

I am asking this question for concept clarification :-

How many integral solutions are there to a+b+c=18 when a $\displaystyle \ge$ 1 , b $\displaystyle \ge$ 2 , c $\displaystyle \ge$ 3 ?

Solution: Let u $\displaystyle \ge$ 0, v $\displaystyle \ge$ 0, w $\displaystyle \ge$ 0, then

a $\displaystyle \ge$ u+1 , b $\displaystyle \ge$ v+2 , c$\displaystyle \ge$ w+3,

Therefore, a + b + c = 18

or u+1 + v+2 + w+3 = 18

or u + v + w =12. From there we solve as usual.

My question is why are we using: Let u $\displaystyle \ge$ 0, v $\displaystyle \ge$ 0, w $\displaystyle \ge$ 0, then

a $\displaystyle \ge$ u+1 , b $\displaystyle \ge$ v+2 , c$\displaystyle \ge$ w+3

Is it to convert each variable i.e a,b and c to one unit each ( since a,b ,c are unequal) or for any other reason ? What is the underlying logic ? Please advise on the above.

2. ## Re: Concept Clarification

What is the concept underlying the approach ?

3. ## Re: Concept Clarification

It is more simple to deal with variables when they all have the same properties (what they are greater than or equal to). Also if you wanted to do this is a non brute-force way to avoid checking for all solutions then you might get some summation formulas which are often well known for sums going from 0 to n so the change of variables to be greater than or equal to zero might make the sums start from 0 instead of 1,2 or 3

4. ## Re: Concept Clarification

Okay ! what you are telling me is this: the approach given makes all the variables uniform so that they can assume the same range of values e.g 0 to 12 in the given question.

I have one similar question :

The number of non-negative integral solutions of x1+x2+x3+x4$\displaystyle \le$ n ( where n is a positive integer ) is :

Solution: Let x5 be such that x1+x2+x3+x4+x5=n

Then required number of solutions= n+5-1C5-1

So, what is the role of x5 here ? Does it convert the inequality to equality ? How does this makes the variables uniform ?

Hi Shakarri,

6. ## Re: Concept Clarification

Originally Posted by SheekhKebab
I am asking this question for concept clarification :-
How many integral solutions are there to a+b+c=18 when a $\displaystyle \ge$ 1 , b $\displaystyle \ge$ 2 , c $\displaystyle \ge$ 3 ?
Having taught this for many years, I will say that I have never found the given approach useful to students. I found it was generally confusing to students.

I would ask them to think of 18 ones being placed into three cells.
If we go ahead and place 1 ball into a, 2 balls into b and then 3 balls into c, then we have 13 balls left to go into a,b, or c: $\displaystyle \binom{13+3-1}{13}$.

Solve in the positive integers with values at least three: $\displaystyle p+q+r+s+t=35$,
ans. $\displaystyle \binom{20+5-1}{20}$

7. ## Re: Concept Clarification

Originally Posted by SheekhKebab
I have one similar question :

The number of non-negative integral solutions of x1+x2+x3+x4$\displaystyle \le$ n ( where n is a positive integer ) is :

Solution: Let x5 be such that x1+x2+x3+x4+x5=n

Then required number of solutions= n+5-1C5-1

So, what is the role of x5 here ? Does it convert the inequality to equality ? How does this makes the variables uniform ?
Hi Plato ,

You missed the question. I was asking a separate question which is mentioned above and which I posted before in the same thread. I was asking about x5 ​ as given above and it's role . Check out the thread.