When three students, Al, Betty, and Charlie, compete in a competition, there are 13 possible orders of finish, allowing for the possibility of ties. These possibilities are illustrated in the chart below:

| First | A | A | B | B | $\mathrm{C}$ | $\mathrm{C}$ | $\mathrm{ABC}$ | $\mathrm{AB}$ | $\mathrm{AC}$ | $\mathrm{BC}$ | A | B | $\mathrm{C}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Second | B | $\mathrm{C}$ | A | $\mathrm{C}$ | A | B |  | C | B | A | $\mathrm{BC}$ | $\mathrm{AC}$ | $\mathrm{AB}$ |
| Third | C | B | $\mathrm{C}$ | A | B | A |  |  |  |  |  |  |  |

When four students, David, Erin, Frank, and Greg, compete in a competition, how many possible orders of finish are there, allowing for the possibility of ties?
There are several possibilities for the groupings of students at the various positions:

 (i) 4 students tied

 (ii) 3 students tied plus 1 student separately

 (iii) 2 students tied plus 2 students tied

 (iv) 2 students tied plus 1 student separately plus 1 student separately

 (v) 1 student separately plus 1 student separately plus 1 student separately plus 1 student separately

In the following table, we summarize the information about each case, the number of ways of arranging the finish of the groups ( $a$ ), the number of ways of selecting the students to be placed in the groups $(b)$, and the total number of possible orders in this case $(a b)$ :

| Case | $a$ | $b$ | $a b$ |
| :---: | :---: | :---: | :---: |
| (i) 4 | 1 | 1 | 1 |
| (ii) $3+1$ | 2 | 4 | 8 |
| (iii) $2+2$ | 1 | 6 | 6 |
| (iv) $2+1+1$ | 3 | 12 | 36 |
| (v) $1+1+1+1$ | 1 | 24 | 24 |

We explain the calculations for two of the cases: (ii) and (iv).

In case (ii), there are 2 possible orderings for the groups of student (group of 3 followed by group of 1 , or group of 1 followed by group of 3 ) and there are 4 ways to place the students in groups (we choose 1 of 4 to be the group of 1 , and the remaining 3 are automatically placed). This gives $2 \times 4=8$ orderings overall.

In case (iv), there are 3 possible orderings for the groups ( 2 then 1 then 1 , or 1 then 2 then 1 , or 1 then 1 then 2 ) and 12 ways to place the student (there are 4 choices for the student in the first group of 1 , for each of which there are 3 choices for the student in the next group of 1 , and then the remaining students are placed in the group of 2 ). This gives $3 \times 12=36$ orderings overall.

In total, there are $1+8+6+36+24=75$ possible orders of finish.

ANSWER: 75