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Problem Description

A championship 18 hole golf course has holes with a par scores of 3, 4, or 5 strokes and a total par for the course of 72 strokes.

So how many different combinations of holes could there be that meet these conditions? (For extra credit, how many arrangements of each of the possible combinations?)

Background & Techniques

Here's a little program that should provide some  practice working with permutations and combinations.    We need to select triples of hole counts (a, b, c) of par 3, 4 and 5 holes that satisfy the equations: a + b + c =18 and 3a + 4b + 5c = 72.  We'll accomplish this  by trying all possible combinations in a nested loop letting a run from 0 to 18 while b runs from 0 to 18-a.   For each a and b, set c = 18-a-b and see if the total of the par scores for all 18 holes is 72.   If so, we have a solution.  

The number of possible arrangements of holes for each a, b, c set is a little more complicated.  Assume that we need to select all possible arrangements of sets of two par 3 holes.  My first thought was that we have 18 choices for the first selection and 17  choices for the second, or  18 X 17 =  306 ways total.  This is the number of permutations of 2 things selected from 18 . 

But ... notice that the 306 permutations includes the pairs (hole1, hole2) and (hole2, hole1) for any two holes.  These shouldn't count twice since the order in  which we selected the par 3 holes doesn't matter to the player (he'll play from 1 through 18 regardless of how we assigned the par values to the holes).  The phrase "order doesn't matter" is our clue that we actually need the number of combinations, the unique ways  we can select 2 things from 18.  The number of combinations is the number of permutations with the duplicates removed.  How many duplicates are there?  For each pair (x, y) selected in our 306 choices there is a corresponding pair (y, x), so we need to divide the number of permutations by 2.  If we were selecting 3 holes, there would be 6 duplicates of each triplet selected (x, y ,z), (x, z, y), (y, x, z), (y, z, x), (z, x, y), (z, y, x) so we would have to divide the permutation count by 6.   In general when selecting R things there will be R! duplicates.   Since the number of permutations is N! / (N-R)!,  the number of combinations is sometimes formulated as (N!) / ((N-R)! x R!). 

In the Permutes and Combinations functions in this program that we won't use the factorial formulas - instead we'll just multiply or divide by the index of a loop as necessary to compute the result.  

So far we've only discussed selecting the number of unique arrangements for one of the hole sizes.  Say we need a par 3's,  the number is given by Combinations(a,18).  To complete the calculation we need to select say b par 4's from the remaining 18-a holes.  This is given by Combinations(b,18-a).  The c  par 5's can only be selected in one way, since c  is 18-a-b by definition and there are only 18-a-b holes remaining.    The total ways we can select the par 3's and 4's is the product of the two combination values.   Answers should be the same of course, no matter which two hole sizes are used for these calculations. 

Running/Exploring the Program 

Suggestions for Further Explorations

Modified:  May 15, 2018

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