Henry Ernest Dudeney/Puzzles and Curious Problems/206 - The Triangular Plantation/Solution

by : $206$

 * The Triangular Plantation

Solution

 * $1216$

Proof
The number of ways to select $3$ from $21$ is the binomial coefficient $\dbinom {21} 3$ which evaluates to $\dfrac {21 \times 20 \times 19} {3 \times 2 \times 1} = 1330$.

All of these form a triangle except where the trees are in a straight line.

Hence we need to eliminate all the latter.


 * Dudeney-Puzzles-and-Curious-Problems-206-solution.png

Observe that the triangle of $21$ consists of lines of $1$, $2$, $3$, $4$, $5$ and $6$, which can be constructed in $3$ different ways, depending on which corner you place the $1$ at.

For each line of $6$, there are $\dbinom 6 3 = 20$ lines of $3$.

In each line of $5$, there are $\dbinom 5 3 = 10$ lines of $3$.

In each line of $4$, there are $\dbinom 4 3 = 4$ lines of $3$.

In each line of $3$, there is just the one line of $3$.

Then there are the $3$ vertical lines of $3$, each of which is one more straight line to be eliminated

This gives $20 + 10 + 4 + 1 + 3 = 38$ arrangements to be eliminated from the count.

Each of these $38$ straight lines are found in each of the $3$ orientations of the triangle of $21$ trees.

Hence the number of lines of $3$ is $3 \times 38 = 114$

So the total count of triangular arrangements is $1330 - 114 = 1216$.