HELP!!!MATHS!!!
posted by ianian .
A video rental store offers 436 different movies for rent. The movies are categorized as Comedies, Action Movies, and Dramas. A movie may be in more than one of the categories. If 234 movies are categorized as Comedies, 97 as Action Movies, and 191 as Dramas, what's the most number of movies that could be categorized as all three?

N(A or C or D) = N(A) + N(C) + N(D)  N(A and C)  N(A and D)  N(C and D) + N(A and C and D)
436 = 97 + 234 + 191  N(A and C)  N(A and D)  N(C and D) + N(A and C and D)
I feel we need more information here 
:'(

First find the surplus, i.e. (234+97+191)436=86.
Now if you want to maximise the 'all three' section, you have to distribute these surplus 86 into this section.But if you add 1 to this region,you are taking care of surplus of 2. Similarly if 10 is added, surplus of 20 gets taken care of. Now for distributing 86 surplus, we have to add 43 to this region, hence the answer.
soln:43
Similarly, the question can be asked to find out minimum number in the 'all three' region. In that case 86 surplus must be distributed in the region of intersection of 2 circles. Now adding 1 in this region takes care of surplus 1, so adding 86 will take care of 86 surplus. So in this case 0 would be the answer for minimum number for 'all 3' section. 
Does anyone here want a d!ck in their pu$$y plz be honest and if you want It i will give it to you! Seriously.