The difficulty in this problem is that there is some overlap in the various groups given. For example the 24 people that have a hot dog includes the three that have a hot dog, soda, and a Popsicle. To sort this out we contruct the matrix below with columns H=Hot dog, S=Soda, and P=Popsicle. There is a row for each combination of items a person could have. For instance, the fourth row indicates people that had a soda and a Popsicle. The Sn columns indicate stages in the solution where the unknowns (indicated with a question mark - ?) are systematically eliminated.
| H | S | P | Given | S1 | S2 | S3 | S4 |
|---|---|---|---|---|---|---|---|
| 0 | 0 | 0 | ? | ? | ? | ? | 15 |
| 0 | 0 | 1 | 38 | ? | ? | 23 | 23 |
| 0 | 1 | 0 | 33 | ? | ? | 21 | 21 |
| 0 | 1 | 1 | 10 | ? | 7 | 7 | 7 |
| 1 | 0 | 0 | 24 | ? | ? | 14 | 14 |
| 1 | 0 | 1 | 8 | ? | 5 | 5 | 5 |
| 1 | 1 | 0 | 5 | ? | 2 | 2 | 2 |
| 1 | 1 | 1 | 3 | 3 | 3 | 3 | 3 |
For the first stage (S1) we know there are exactly thee people with a hot dog, a soda and a Popsicle, so we mark 3 in red. Moving to the next stage (S2), we are given that there are five people with a hot dog and a soda (second to last row). Since the three folks with a hot dog, a soda, and a Popsicle have already been accounted for in this group, that leaves two individuals that had only a hot dog and a soda. We indicate this fact with a red 2 in the S2 column.
Continuing in this fashion, we find in the last stage that there were 15 people that had nothing.