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?

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- Oct 27th 2012, 07:43 AMPetrusWords problem nr2
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?

- Oct 27th 2012, 08:00 AMShakarriRe: Words problem nr2
Call all the people in analysis group A and all those in discrete are group B

The number of people in A and B is AUB

The number of people not in both is A + B - AUB = 361 + 223 - 188= 396 - Oct 27th 2012, 08:09 AMPetrusRe: Words problem nr2
thats so logic just ohh well dont know why i thought like u :s

- Oct 27th 2012, 08:27 AMSorobanRe: Words problem nr2
Hello, Petrus!

There is just barely enough information.

Quote:

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: .$\displaystyle n(A \vee D) \:=\:188 + 173 + 35 \:=\:396$