Minimum number of students possible . In a business school there are 3 electives and at least one elective is compulsory to opt. 75% of the students opted for marketing, 62% opted for finance and 88% opted for H.R. A student can have dual or triple specialization.
What is the minimum number of students that specialize in all three streams ?
Can someone please help?
 A: $0.75+0.62+0.88 = 2.25$
The average student is taking $2.25$ electives.
If $x_1$ is the # of students taking one elective,
$x_2$ is the # of students taking two electives, and 
$x_3$ is the # of students taking 3 electives.
$x_1 + 2x_2 + 3x_3 = 2.25\\
x_1 + x_2 + x_3 = 1$
To minimize $x_3,$ we must assume that $x_1 = 0$
$2x_2+ 3x_3 = 2.25\\
x_2 + x_3 = 1$
$x_3 = 0.25$
A: By Inclusion-Exclusion we have that:
$$|M \cup F \cup H| = |M| + |F| + |H| - |M \cap F| - |M \cap H| - |F \cap H| + |H \cap F \cap H|$$
and thus:
$$|M \cap F \cap H| = |M| + |F| + |H| - |M \cap F| - |M \cap H| - |F \cap H| + |H \cup F \cup H|$$
Now, we know that $|M| = 75$, $|F| = 62$, $|H|=88$, and $|H \cup F \cup H|=100$
So, to minimize $|M \cap F \cap H|$, we need to maximize $|M \cap F|+|M \cap H|+|F \cap H|$. Or, in other words, we want as many students as possible to take two subjects, but not all three.
Now let's look at a Venn Diagram:

OK, so we figured that in order to minimize the number of students in $IV$, we want to maximize the number of students in regions $I$, $II$, and $III$
Well, for region $I$, since there are $88$ students doing H.R, we can at most have $12$ students not doing H.R, and so at most $12$ in region $i$. so, let's put $12$ in region $I$. Similarly, with $62$ students doing Finance, we can at most have $38$ students in region $II$. And finally, with $75$ students doing Marketing, we can at most have $25$ students in region $III$. So, that gives $12+38+25=75$ students in regions $I$, $II$, and $III$. And since we didn't put anyone in a region denoting students taking a single subject, that means there are $100-75=25$ students left for region $IV$. So that's the minimum: $25$
Does this method always work? No. This time we were able to have all students take multiple streams. But, suppose the numbers were as follows:
$|M| = 25$, $|F| = 52$, $|H|=38$, and again $|H \cup F \cup H|=100$
Then using the above method we would have tried to put $62$ students in region $I$, $48$ in region $II$, and $75$ in region $III$, but $62+48+75=185>100$, so that is impossible. However, when this happens, we can simply fill up the regions until we get the total number of students, e.g. we could put $25$ in region $I$, $27$ in region $II$, and $0$ in region $III$, and while this means that $11$ students will be taking H.R. as their single subject, there are $0$ students taking all $3$ subjects.
A: Suppose students 1 through 75 do marketing, 39 through 100 do finance, and 1 through 38 and also 76 through 100 and also some others do HR, whatever that is. Well, that accounts for $38+25=63$ of the 88 students doing HR, so there must be another 25 doing all three subjects. 
A: Call the set of students specialising in marketing $M$, the set of students specialising in HR $H$ and the set of students specialising in finance $F$.. The least number of students that can specialise in both marketing and finance is $n_{MF} = |M| - (|M \cup F \cup H| - |F|) = 75 - (100 - 62) = 37$.
The number of these students that do not also specialise in HR is then at most $n_{MF} - (|M \cup F \cup H| - |H|) = 12$.
So there are at least $25$ students with all three specialisations.
A: Let $S$ be total number of students; $x_1,x_2,x_3$ are taking only $M,F,H$; $y_1,y_2,y_3$ are taking $M\cap F \setminus H,M \setminus cap H \setminus F,F\cap H \setminus M$; $z$ are taking $M\cap F\cap H$. Then:
$$x_1+y_1+y_2+z=0.75S \\ x_2+y_1+y_3+z=0.62S \\ x_3+y_2+y_3+z=0.88S.$$
To be whole numbers $S$ must be a multiple of $100$. If $S=100$, then by adding the three equations we get:
$$z=x_1+x_2+x_3+25.$$
To make $z$ minimum $25$, we must set $x_1=x_2=x_3=0$. 
Hence:
$$y_1=38,y_2=12,y_3=25.$$
If $S=200$, then by repeating the above calculations we find:
$$x_1=x_2=x_3=0; z=50; y_1=138,y_2=112,y_3=125,$$
which is not suitable.
Thus $\min  z=25$.
