Consider the equation x+y+z=18. Here, x,y and z are whole numbers. To find the number of solutions of this equation, i would perhaps use what's famously known as the Stars and Bars method. But, this method is only applicable when there is no restriction on values that x,y and z can take. However, if I add an extra condition that each of x,y and z can only take on integral values that belong to the closed interval [1,8], then this method is not applicable. Our professor prescribes the use of "coefficient method" for such a situation. In this case, we would obtain our answer by finding the binomial coefficient of x^18 in the expansion of (x¹+x²+x³+x⁴+...x^8)^3. I notice here that the polynomial only contains powers of x which are present in the interval [1,8]. Also the polynomial is raised to a power 3, which represents the number of variables involved ( in this case , x,y and z). I've understood how to apply this method to solve different problems. I'm fascinated by the fact that binomial and multinomial theorems are coming in handy in permutations and combinations. But, I haven't understood the "coefficient" method I've explained above, at all. Why does it work? I don't get the link.
Asked
Active
Viewed 72 times
1
-
It is not that stars and bars can't be used, with inclusion-exclusion as may be required. You can have a look at the linked answer which explains how stars and bars could be used in such a situation, and also gives an elementary understanding of your professor's "coefficient method" (termed generating functions) See if your doubts get addressed. https://math.stackexchange.com/questions/2007216/how-many-different-possible-permutations-are-there-with-k-digits-that-add-up-to/2007514#2007514 – true blue anil Sep 07 '24 at 21:39
-
To further elucidate, the particular type of generating function used here is called am ordinary generating function, the coefficients of x in each term represent how many times that term occurs, and as you know, when you multiply polynomials, the coefficients of the terms get added, so you can find out the number of ways to get the desired sum. You can read up generating functions, or look up further examples on this site. – true blue anil Sep 07 '24 at 21:59
-
I agree with both of the comments above. As a further example of using Stars and Bars, combined with Inclusion-Exclusion, to solve the problem, see this answer. – user2661923 Sep 07 '24 at 22:08
1 Answers
1
As a bonus, I am posting a very simple answer using stars and bars with a little inventiveness.
First fill $1$ each in the $3$ "bins", we now need a sum of $15$
Assume that the now maximum of $7$ possible in each bin have been put in, totalling $21$
Now take out $6$ using stars and bars, thus
$\dbinom{6+3-1}{3-1} = \dbinom82 = 28$
true blue anil
- 48,250