What is the concept underlying the approach ?
I am asking this question for concept clarification :-
How many integral solutions are there to a+b+c=18 when a 1 , b 2 , c 3 ?
Solution: Let u 0, v 0, w 0, then
a u+1 , b v+2 , c 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 0, v 0, w 0, then
a u+1 , b v+2 , c 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.
Thanks in advance !
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
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 x_{1}+x_{2}+x_{3}+x_{4} n ( where n is a positive integer ) is :
Solution: Let x_{5} be such that x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=n
Then required number of solutions= ^{n+5-1}C_{5-1}
So, what is the role of x_{5} here ? Does it convert the inequality to equality ? How does this makes the variables uniform ?
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: .
Solve in the positive integers with values at least three: ,
ans.