Overlapping Sets :

[vikramsumer]

Got It


I did this question by the Venn diagram method and made an error in the calculation. I need to add a+b+c+3 to the left hand side of my equation and I get the correct answer 10.

Thanks

[tim]

cool; glad to hear it..

[mneeti]

I tried solving the question this way and got the solution as well but when i try to distribute the numbers I get confused. Please suggest where I am going wrong :

Solution -

3 are registered for all classes.
So subtract 3 from all.
History=22,Maths=22 & English = 31
Addition becomes 22+22+31 = 75
Out of 68 we already have discarded 3 students, present in all three
Remaining = 65.
But for 65 students addition is coming to 75 means there should be 10 students taking exactly 2 subjects.
So ans is 10 which is B

Confusion:

Now when I try to distribute the numbers I do not know what I am doing :/

we have 68 students in total which includes 3 students in all the 3 subjects, 10 with exactly 2 subjects and balance is exactly 1 subject

68 = 3+10+55 ? According to this, the number of students with exactly one subject is 55 but in my solution above, I have reduced 65 from 75 assuming that these are the students with exactly one subject.

I am not able to understand this. Please help. Thanks !

[RonPurewal]

3 are registered for all classes.
So subtract 3 froom all.
History=22,Maths=22 & English = 31
Addition becomes 22+22+31 = 75
Out of 68 we already have discarded 3 students, present in all three
Remaining = 65.

the problem is here -- you appear to be thinking that this figure actually represents 65 different people.
it doesn't, because you haven't yet accounted for the students who are taking two subjects. yes, you've eliminated the three-subject students, but each person taking two subjects is counted twice in this figure of 65.
since there are ten students taking 2 classes each, those ten students are counted twice. therefore, this figure only represents 55 different people.

[buymovieposters]

for these 3 overlapping set problems do you always want subtract abc 3 times? will the 3 always be included?

# of items total = (a + b + c) - (ab + ac + bc) + 3(abc)

one extremely efficient way to attack this problem is to use the following formula, which works for all combinations of three sets a, b, c:

# of items total = (a + b + c) - (ab + ac + bc) + (abc)

in the formula, 'a' stands for the # of items in set a (regardless of whether they are in other sets as well), 'ab' for the # of items that are in both sets a and b (again, regardless of whether they're in c), and so forth.

note that you cannot use this formula if you are given information with exclusivity (for instance, the number of items in set a and b but not in set c).

in this problem, you are looking for the value of (ab + ac + bc - 3abc). notice that you have to subtract abc 3 times, because those items have been tallied three times (they are in 'ab', they are in 'ac', and they are also in 'bc'). since you already have abc = 3, all you need is the value of (ab + ac + bc).

using the formula,
68 = (25 + 25 + 34) - (ab + ac + bc) + 3
(ab + ac + bc) = 19
answer = 19 - 3(3) = 10

you can also solve this problem with venn diagrams and associated reasoning. hit up this thread again if you want more details about that.

[geezer0305]

Hi,

Shouldnt the answer be 13?

Solution:


total number of students = (Total in English)+(Total in History)+(Total in Math)- (all students with 2 common subjects) - (all students with 3 common subjects)

=> 68 = 25+25+34-(all students with 2 common subjects)-3

=> (all students with 2 common subjects) = 13

Is my reasoning flawed? I solved it using the Venn Diagram.

[jnelson0612]

Hi,

Shouldnt the answer be 13?

Solution:


total number of students = (Total in English)+(Total in History)+(Total in Math)- (all students with 2 common subjects) - (all students with 3 common subjects)

=> 68 = 25+25+34-(all students with 2 common subjects)-3

=> (all students with 2 common subjects) = 13

Is my reasoning flawed? I solved it using the Venn Diagram.

Take a look at Ron's first and second posts on the first page of this thread. :-)

[RonPurewal]

for these 3 overlapping set problems do you always want subtract abc 3 times? will the 3 always be included?

# of items total = (a + b + c) - (ab + ac + bc) + 3(abc)

that doesn't match what i posted, so, no.

the best way to check this for yourself is to actually make concrete lists/rosters of some sort, and then check whether your formula works.
e.g., write up some lists of people who play basketball, football, and/or baseball, with some players playing 2 of the sports and a few playing all three. (or, just draw a giant venn diagram and write the names directly in the diagram).
then, just count things up and see which formulas work and which don't.

[fzaffer]

Could you please show me the Venn Diagram way? Just for my personal understanding.

Would appreciate it.

Thank you.

[RonPurewal]

* draw three overlapping circles.

* the number in the middle region (the region that's common to all three circles) is '3'.

* at this point you're going to need to do one of two things:
1/ define annoyingly many variables,
2/ pick numbers.

each of the three circles is going to be divided into four parts (the middle part, the two parts shared with one other circle each, and the one part that belongs to just that one circle by itself).

if you're going to pick numbers:
pick one of the circles.
you'll have the '3' in the middle zone...
... but then you can just pick numbers for the other three parts, so that the whole circle has the specified total.
then fill in numbers for another circle.
once you've done that, there should be only one region remaining (the one belonging only to the third circle). you'll be able to use subtraction to figure out the number that goes there.
(if you end up with any negative numbers, pick other numbers and do the process again.)

[fzaffer]

Thank you Ron!

[tim]

[RonPurewal]

you're welcome.

[NNK314]

you're welcome.

Hi Ron/ Tim,

I understand how to use the 3 solving methods for overlapping sets (the both/neither formula, the double set matrix and the venn diagram). My question is, when you look at a problem, is there a way to just know which method will work best for the problem?
For example, after setting up the DSM, and then realizing half way that it might not be the best approach wont leave me with any time to use the other approaches and I would have to bail on a problem that I could/ should have solved using another method.....?
Or are the 3 methods interchangeable and either one works for any problem... so basically if I'm not able to solve the problem using one method, I've probably already made a mistake in comprehension of the problem and should just bail and move on...

Thanks in advance!

[RonPurewal]

you just pick a strategy, throw it at the problem, and see what happens.

For example, after setting up the DSM, and then realizing half way that it might not be the best approach wont leave me with any time to use the other approaches and I would have to bail on a problem that I could/ should have solved using another method.....?

that red thing isn't a thing.

if you are actually honest with yourself about whether you're making genuine progress on the problems—and you STOP THE VERY MOMENT you cease making progress—then you will ALWAYS have time to try as many methods as you need to try.