is this the correct way to solve the question

Question

if there are 6 runners and 3 high jumpers but 2 are also high jumpers and runners,how many people are on the athletic team

my work

6 runner +3 HIGH JUMPERS=9-2=7 members on the athletic team

Answer 1

Some people are runners. Some people are jumpers. Some people are both. Some are neither.

Everyone on the team is one or the other or both. Noboby on the team is neither.

6 of the team are runners. Some of them might or might not be jumpers as well.

3 of the team are jumpers. Some of the might or might not be runners as well.

2 of the team are both (which means the rest are one or the other and not both).

So how many members are there on the team?

Answer 2

Think about your answer for a moment. The team consists of $6$ runners and $3$ high jumpers, so even if every member of the team competed in just one of the two events, the team would have only $6+3=9$ members. It cannot possibly have $36$ members: that answer simply isn’t even reasonable.

Perhaps it would help to draw a picture. We have $6$ runners, whom I’ll represent by $6$ $R$s, and $3$ high jumpers, whom I’ll represent by $3$ $H$s:

$$R\,R\,R\,R\,R\,R\,H\,H\,H$$

Oops! Two of those high jumpers are also runners; I’ll represent each of them by $(RH)$.

$$R\,R\,R\,R\,(RH)\,(RH)\,H$$

Or you could start with this picture:

          R R R R R R H H H

and rearrange it like this:

          R R R R R R
                  H H H

to show that two of the high jumpers are also runners. Either way it should now be clear that the team really has only $7$ members.

But we can arrive at this correct result without resorting to pictures. When we initially compute a total of $6+3=9$ members, we’re counting every athlete who competes in both events twice, once as a runner and once as a high jumper. This means that we’ve overcounted by $2$, and our provisional total of $9$ is too large by $2$; to correct it, we subtract the overcounting and find that the team actually has just $9-2=7$ members.

Multiplying the figures is for a completely different kind of problem. If we had $6$ athletes who specialized in running events, $3$ who specialized in the high jump, and $2$ who did both, and we wanted to form a team of $3$ consisting of one specialized runner, one specialized high jumper, and one athlete who could do both, there would be $6\cdot 3\cdot 2=36$ possible different teams of $3$. This is an application of the product rule for computing the number of ways to make several independent choices. In the your problem here we’re not making choices: we’re just counting the athletes and trying to make sure that we count each one exactly once.