SOLUTION: Will Someone Help Me With This Problem? In a survey of four-year colleges and universities, it was found that: 356 offered a liberal arts degree. 293 offered a computer engi

Let L = the set of all offering liberal arts, regardless
of whether they offered the other two degrees.
Let C = the set of all offering computer engineering, regardless
of whether they offered the other two degrees.
Let N = the set of all offering nursing, regardless
of whether they offered the other two degrees.

There are 8 regions which I have labeled with arbitrary small
letters.



The most inclusive clue we have is this one:

Those 68 are in all three circles, and the only
region that is common to all three circles is
the middle region f, so we put 68 for f



Now we look at this:

68 of those 193 are already accounted for in the middle region,
so that leaves 193-68 or 125 to go in the other region common
to circles L and C, which is region e, so we put 125 in that region:



Now we look at this:

68 of those 200 are already accounted for in the middle region,
so that leaves 200-68 or 132 to go in the other region common
to circles L and N, which is region h, so we put 132 in that region:



Now we look at this:

We have numbers in 3 of the regions of circle L, so 
132+68+125 or 325 of those 356 are already accounted
for so that leaves 356-325 or 31 to go in the remaining
region of circle L, region d, so we put 31 in that region:



Now we have completed all four regions of circle L,
and see that if we add up all four regions we have
31+125+68+132 we have 356 in circle L.

Now we look at this:

68 of those 139 are already accounted for in the middle region,
so that leaves 139-68 or 71 to go in the other region common
to circles C and N, which is region j, so we put 71 in that region:



Now we look at this:

We have numbers in 3 of the regions of circle C, so 
125+68+71 or 264 of those 293 are already accounted
for so that leaves 293-264 or 29 to go in the remaining
region of circle C, region i, so we put 29 in that region:



Now we look at this:

We have numbers in 3 of the regions of circle N, so 
132+68+71 or 271 of those 285 are already accounted
for so that leaves 285-271 or 14 to go in the remaining
region of circle N, region k, so we put 14 in that region:




The most EXcluding clue we have is this one:

Those 26 are outside all three circles, and the only
remaining region is outside all three circles,
the region m, so we put 26 in region m, and our Venn 
diagram is now complete:



So we are now ready to look at the questions, for now
we can answer anything we are asked because we have
the complete Venn diagram:

We add up all the numbers in all 8 regions:

31+125+29+132+68+71+14+26 = 496

That comes from the three regions which are parts of circles L and N,
which are not parts of circle C.  Those are the three circles
originally labeled d,h, and k, which now contain the numbers
31,132, and 14, and when we add 31+132+14 we get 177.  

That is the one region that is withing circle C which is not part
of either of the other two circles.  That is the upper right region
that was originally labeled i, which now contains 29.  

That would be the three regions originally labeled e,h, and j,
which are only parts of two circles, so the answer here is
125+132+71 or 328

That is all of the regions which are parts of circles.  The easy way
to find it is to take the answer to (a) which was 496 and subtract 
the 26 that are not in any of the circles and get 496-26 or 470.

Edwin