Question 1166661
<font face="Times New Roman" size="+2">


There are 100 students, 30 do not major in any of the three disciplines, so there are 70 that majored in at least one.  Note that 33 who majored in Computer Engineering counts the students that majored in only CE and the CE/CS and CE/M dual majors AND the students that majored in all three.  So if CE is 33 and CS is 30 and M is 33, that is a total of 96 which is 26 more than the 70 total students majoring in at least one.  Hence, 26 are being multiple counted.  Then the total of the three dual majors (which double counts the three major people is 11 plus 12 plus 13 is 36.  36 minus 26 is 10 and that is the number of triple major people.


Then CE/CS minus 10 is 1, so there is exactly 1 CE/CS only major.  Likewise, there are exactly 2 CE/M only majors, and exactly 3 CS/M only majors.


33 minus 10 triples, 1 CE/CS, and 2 CE/M leaves 20 CE only majors.  Then 30 minus 10, 1, and 3 leaves 16 CS only majors.  And 33 minus 10, 2, and 3 leaves 18 M only majors.  20 plus 16 plus 18 = 54 single majors.

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">


I > Ø
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
</font>