Action Dram-Coms

let a, b and c be movies that are comedies, action and dramas only, respectively. let d,e and f be movies that are comedies and action, action and drama, and drama and comedy, respectively. let g be movies that are all 3. as we can see, a+b+c+d+e+f+g=436---(1). However, we also see that a+d+e+g= 234 ---(2)since they are all movies that are comedies, b+e+f+g= 97 ---(3)and c+f+d+g=191---(4) due to similar reasons. add (2),(3) and (4) and we get a+b+c+2d+2e+2f+3g=522---(5). subtract (1) from (5) and we get d+e+f+2g=86. Therefore for g to be maximum, d+e+f must be 0 and thus we see that 2g=86 and the maximum value of g= 43

This solution is long but very explanatory. Here we go.

Firstly, note a very important thing. Since we want to find the most number of movies that could be categorized as of all three genres , we must assume that all the overlapping occurs in triples, and not in doubles, that is, a movie can be of all three categories, but it cannot be of only two of them. Secondly, for those who don't know, genre is the category of movies. Here, the number of genres is $3$ .

So, let's assume the number of movies that could be categorized as all three genres be $x$ . So, in all three genres, there are $x$ movies which are a part of other two genres as well. Therefore, the number of movies specifically falling in a particular genre is given by: Total movies in the genre minus Movies in all three genres .

Therefore, in each genre, number of movies specifically falling in it are: $Comedies: 234-x$ $Actions: 97-x$ $Dramas: 191- x$

Now, it is clear that the sum of movies of all genres (specifically) and the movies present in all three genres should be equal to the total number of books. Plugging them together, we get: $234-x+97-x+191-x + x = 436$ Simplifying it yields us $x = \fbox{43}$ .

Therefore, the most number of movies that could be categorized as all three movies is $\fbox{43}$ .

Let set A denote comedy movies, $|A|=234$ . Let set B denote action movies, $|B| =97$ . Let set C denote drama movie, $|C| =191$ . We are given that $|A \cup B \cup C| = 436). Set $|A \cap B \cap C| = x$ , which we wish to maximize.

From the Principle of Inclusion and Exclusion, we have $| A| + |B| + |C| - |A\cap B| - |B\cap C| - |C\cap A| + |A\cap B \cap C| = 436$ . Since $A \cap B \cap C \subset A\cap B$ , it follows that $|A\cap B \cap C| \leq |A \cap B|$ . Similarly, we have $|A\cap B \cap C| \leq |B \cap C|$ , $|A\cap B \cap C| \leq |C \cap A|$ . Hence, the equation at the start of the paragraph gives us $234 + 97 +191 - x - x - x + x \geq 436$ , or that $x \leq 43$ .

[Latex edits. Edits for clarity. - Calvin]

First find the surplus, i.e. (234+97+191)-436=86.

Now if you want to MAXIMIZE 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.

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 86 would be the answer.

Don't get the last paragraph wrong because it's just showing another solution to find the MINIMUM number of movies that could be categorized as all three and so, the final answer would be 43.

We can either count the movies by listing them all out and counting to get 436, or manipulate the number of movies in each category to get the same sum. But if we just sum 234, 97, and 191, then we could movies in exactly two categories twice, and movies in all three categories thrice! So we need to correct for overcounting. Let $\alpha$ , $\beta$ , and $\gamma$ be the number of movies that are in Comedies and Action Movies; Comedies and Dramas; and Action Movies and Dramas; respectively. These numbers all include the movies that are in all three categories. So we can take the sum:

$234+97+191-\alpha-\beta-\gamma$

This counts the number of movies in exactly one category exactly once, as well as movies in exactly two categories exactly once. But movies in all three categories are counted zero times! If we let the number of movies in all three categories be $x$ , then we have that

$436=234+97+191-\alpha-\beta-\gamma+x$

Thus

$\alpha+\beta+\gamma-x=86$

However, note that $\alpha$ , $\beta$ , and $\gamma$ are by definition at least $x$ , thus $\alpha+\beta+\gamma-x\geq 2x$ , with inequality when $\alpha=\beta=\gamma=x$ . Thus the minimum value is when $x=86/2=\boxed{43}$ . We can use the above work to construct an equality case.

Draw a Venn Diagram. We wish to maximize the number of movies found in all three spaces, therefore we will try to have no movies that are two types but not three (e.g. action and drama but not comedy). We let there be $x$ movies which are categorized as all three, $234-x$ only comedies, $97-x$ only action movies, and $191-x$ only dramas.

Total we have $x + (234-x) + (97-x) + (191-x) = 522-2x$ movies. We set this equal to the total number of movies to get $x$ .

$522-2x=436$

$x = 43$

Luckily this is an integer and everything works out. If it had not, we would have had to play with some other values.

Here we want to maximize the number of movies in all 3 categories.

For maximizing this number we assume that there is no such movie that is in 2 categories.

Suppose, x movies are in all three categories

Therefore, we can derive the total number of movies by subtracting the common movies from all 3 unique categories and finally adding them for 1 time(otherwise,they will be totally out of business!!!)

Thus the equation looks like the following :

     (234 - x) + (97 - x) + (191 - x) + x = 436
     -2x = -86
      x = 43

therefore, the maximum number of movies in all 3 categories is 43.

Let x be the number of movies which are both comedy and action movies and y be the number of movies which are both comedy and drama. Now, No. of movies which are only action movies = 97-x & No. of movies which are only drama movies = 191-y. now, total no. of different movies = no. of comedy movie + no. of movies which are only drama + no. of movies which are only action movies or, 436 = 234 + 191-y + 97-x or, x + y = 86. since x represents movies which are both comedy and action and y represents movies which are both comedy and drama, the common part of x and y represents the movies which belongs to all categories( say p) for p to be maximum, x = y = 86/2 = 43. which is the required answer.

The maximum possible cds which are common to all the three fields are possible only if there is no cd which is common to only 2 categories.(I) w.k.t. n(A union B union C)=n(A)+n(B)+n(C) - [n(A union B)+n(B union C)+n(C union A)] + n(A intersection B intersection C) Let n(A intersection B intersection C)=x => n(A union B union C)=n(A)+n(B)+n(C) - [n(A union B)+n(B union C)+n(C union A)] +x =>n(A union B union C)=n(A)+n(B)+n(C) - [3x] + x [from(I)] Putting in the values from question=> 436=522 - 2x => - 86 = - 2x =>x = 43 Thus, the number of maximum cds that can be part of all the three given categories are 43.

Let $C, A, D$ be the sets of movies categorized in Comedies, Action Movies and Dramas, respectively. We have $|C| = 234$ , $|A| = 97$ , $|D| = 191$ and $|C \cup A \cup D| = 436$ Now, Principle of Inclusion and Exclusion tells us that $|C \cup A \cup D| = |C| + |A| + |D| - (|C \cap A| + |A + \cap D| + |D \cap A|) + |C \cap A \cap D|$ Also, note that $|C \cap A| = |(C \cap A) - D| - |C \cap A \cap D|$ And similarly for $|A \cap D|$ and $|D \cap A|$ . Therefore $|C \cap A \cap D| = \frac{1}{2} \left( (234 + 97 + 191) - 436 - \sum (|(C \cap A) - D|) \right)$ Thus $|C \cap A \cap D| = \frac{1}{2} \left( (234 + 97 + 191) - 436 \right) = 86$ Equalities hold when any pair of $C, A, D$ intersects only at the intersection of all of them.

If we add them all together we get $234+97+191=522$ so there must be $522-436=86$ repeats- (people who have been counted $twice$ or even $thrice$ ).

The obvious maximum would occur if $ALL$ of these repeats were from overcounting people that appear thrice in total. So if the number of people in all three was $x$ we would have $3x-x=2x$ repeats.

So $2x=86$

$x=\boxed{43}$

The total number of times movies are overcounted is (234+97+191)-436=86. Now if we suppose that all of these overcounted movies are categorized as all three, each movie will be counted three times, and thus overcounted twice. So the total number of 3-category movies is half of 86, which is 43.

In order to maximize the number of movies classified as all three, do not consider any scenarios in which movies have only two classifications.

Let x = number of movies classified as comedy only. Let y = number of movies classified as action only. Let z = number of movies classified as drama only. Let w = number of movies classified as all three.

Write a system of four linear equations in four unknowns.

• $x+w=234$
• $y+w=97$
• $z+w=191$
• $x+y+z+w=436$

Solve for x , y , and z .

• $x=234-w$
• $y=97-w$
• $z=191-w$

Substitute into the last equation from above.

$(234-w)+(97-w)+(191-w)+w=436$

$522-2w=436$

$2w=86$

$w=43$

At most, forty-three movies may be classified as all three.

Let 'a', 'b', 'c' be the Comedy, Action and Drama movies respectively. Let 'd' be Comedy & Drama, 'e' be Comedy & Action and 'f' be Action & Drama. Let 'g' be the all category movies.

Now, according to question,

a + b + c + d + e + f + g = 436, ......................(i)

a + d + e + g = 234, ..................(ii)

b + e + f + g = 97, ..................(iii)

c + d + f + g = 191, .....................(iv)

Then, (ii) + (iii) + (iv) - (i) = 522 - 436

d + e + f + 2g = 86

And for maximizing g we should put d + e + f = 0,

Then, g = 43 (Ans)

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 *

We have 4 variables: Movies that are only Comedies C, that are only Action A, that are only Dramas D and that can be categorized as all three X. Logically, we won't any movie that can be categorized as two types of movie.

So we'll have the next equations:

C+D+A+X=436

C+X=234

D+X=191

A+X=97

Solving them, we'll get the value of X, that is 43.