In a shooting contest, 8 targets are arranged in two columns with 3 targets and one column with 2 targets. The rules are:

- The shooter freely chooses which column to shoot at.
- He must aim at the lowest target not yet hit.

![](https://cdn.mathpix.com/cropped/2024_05_01_4fd1b602786ad9981479g-18.jpg?height=513&width=515&top_left_y=611&top_left_x=652)

a) If the shooter disregards the second rule, in how many ways can he choose just 3 positions out of the 8 distinct targets to shoot at?

b) If the rules are followed, in how many ways can the 8 targets be hit?
Solution

a) If $x, y$ and $z$ are the positions of the targets, in principle, the shooter has 8 choices for $x$, $8-1=7$ for $y$, since we cannot repeat the chosen position, and $8-2=6$ choices for $z$, since we cannot repeat any of the two already selected positions. This gives $8 \cdot 7 \cdot 6$ choices. However, the choices

$$
x y z, x z y, y x z, y z x, z x y, z y x
$$

produce the same set of chosen positions. To avoid repeated counts, we simply divide the previous count by the number of times each set of positions was counted, that is, the total number of choices is

$$
\frac{8 \cdot 7 \cdot 6}{6}=56
$$

In general, the number of ways to choose 3 objects from a set of $n$ distinct objects is

$$
\frac{n(n-1)(n-2)}{6}
$$

b) Suppose the columns are marked with the letters $A, B$, and $C$. Each sequence of shots can be encoded with a linear arrangement of these letters following the order of the shots from left to right, forming a word with 3 letters $A$, 2 letters $B$, and 3 letters $C$. For example, $A A B C C C B A$, means that we will shoot twice in the first column, once in the second, three times in the third, once more in the second, and finally once in the first. The order of the shots in each column is determined by the rules of the competition. Since there is a correspondence between these words and the distribution of shots, it suffices to count the number of distinct words that can be created with these letters. To do this, we will calculate the number of possible anagrams using 3 A's, 2 B's, and 3 C's, in a sequence of 8 letters. Observing the 8 positions, we can choose the 3 for the A's in $\frac{8 \cdot 7 \cdot 6}{6}=56$ ways. Among the five remaining spaces, we must choose 2 for the B's, and this can be done in $\frac{5 \cdot 4}{2}=10$ ways. The three remaining positions will be occupied by the C's. Therefore, there are $56 \cdot 10=560$ ways to hit all the targets with the imposed rules.