# Set Theory, Calendars, Clocks and Binomial Theorem

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## Set Theory: Min and Max % of people

95% of the students in a class have taken Marketing, 80% have chosen Finance, 84% have chosen operations (ops), and 90% have chosen Human Resources (HR). What is the maximum and minimum percentage of people who have chosen all of the four?
1. 80% and 56%
2. 95% and 53%
3. 80% and 49%
4. 80% and 51%

Choice C. 80% and 49%

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## Detailed Solution

If you get the Venn Diagram right, you are halfway through !!

Finding the maximum percentage is easy. If ${F \subset O\subset H \subset M}$ , then the percentage of people who have taken all 4 should be 80% and this is the maximum value it can take.

For the value to be minimum, the numbers should be as far ‘apart´ as possible. Let us do this iteratively. First, let us take Marketing and Finance, and see if we can find the minimum percentage of students who should have taken both.

When M and F, are as far ‘apart´ as possible, ${M\cap F}$ would be minimum. And the minimum value would be 80% + 95% – 100% = 75%.

Now, let us start with this ${M\cup F}$ and add Ops to the mix.

Now, the minimum value of ${M\cap F\cap O}$ would be when these are as far ‘apart’ as possible. And the minimum value would be 75% + 84% – 100% = 59%.

Adding HR also to the mix, we get -

The minimum possible value of ${M\cap F\cap O\cap H}$ = 59% + 90% – 100% = 49%.

As a formula, the minimum value is 100% – (100% – 95%) – (100% – 80%) – (100% – 84%) – (100% – 90%).

= 100% – 5% – 20% – 16% – 10% = 100% – 51% = 49%.

Again, do not go with just the formula. See the trial and error iteration. The trial and error iteration is far more useful than knowing the formula for a template.

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