361 students at a university has taken a course in the analysis, 223 have taken a course in discrete mathematics, and 188 have read both courses. How many students at the university have taken a course in either analysis or discrete mathematics?
361 students at a university has taken a course in the analysis, 223 have taken a course in discrete mathematics, and 188 have read both courses. How many students at the university have taken a course in either analysis or discrete mathematics?
Hello, Petrus!
There is just barely enough information.
361 students at a university has taken a course in Analysis,
223 have taken a course in Discrete Mathematics,
and 188 have taken both courses.
How many students at the university have taken a course in either Analysis or Discrete Mathematics?
We can use this formula:
. . $\displaystyle n(A\,\vee\,D) \,=\,n(A) + n(D) - n(A\,\wedge\,D) $
Or we can place the information into a chart . . .
. . $\displaystyle \begin{array}{c|c|c|c|} & \text{Discrete} & \sim\text{Discr.} & \text{Total} \\ \hline \text{Analysis} & 188 && 361 \\ \hline \sim\text{Analy.} & & & \\ \hline \text{Total} & 223 & & \\ \hline \end{array}$
And we can fill in more data . . .
. . $\displaystyle \begin{array}{c|c|c|c|} & \text{Discrete} & \sim\text{Discr.} & \text{Total} \\ \hline \text{Analysis} & 188 & 173 & 361 \\ \hline \sim\text{Analy.}& 35 & & \\ \hline \text{Total} & 223 & & \\ \hline \end{array}$
Therefore: .$\displaystyle n(A \vee D) \:=\:188 + 173 + 35 \:=\:396$