Chapter Summary: Mixed Solved Problems on Sets

Transcript

0:00Introduction

Hello everyone! Alright, one quick question to warm up. Take the word Allahabad, and make the set of its letters. How many subsets does this set have? Count the letters: nine. So, two to the power nine, five hundred and twelve subsets. Easy, right? Hmm, are you sure? Keep thinking. This is our chapter summary on sets: mixed problems, quick tricks, and the small traps that JEE loves. So, let's begin.

0:32Subsets: count only the distinct elements

So, let's settle our puzzle first. Inside a set, a repeated element counts only once, okay? Allahabad has nine letters, but A and L repeat. So the set of letters, in curly brackets, has just five elements: A, L, H, B and D, nothing more. Five elements, and the double arrow means, implies: two to the power five, thirty two subsets. Not five hundred and twelve! Look, this is a favourite JEE trap. Find the distinct elements first, then use two to the power n. Alright, one more fact. If a set has n elements, it has two to the power n subsets. How many of them have an odd number of elements? Yes, exactly half! Fix one element, and pair each subset with the one you get by putting that element in, or taking it out. The sizes differ by one, so one is odd and the other even. So, for n at least one, exactly half are odd. Clear? So if five hundred and twelve subsets have odd size, what is n? Half of two to the n is two to the power, n minus one. And five hundred and twelve is two to the power nine. So n minus one is nine, and n is ten. See? Our five hundred and twelve is back, but now it counts only the odd sized subsets, of a ten element set. Got it?

2:00The universe decides the answer

Next, an easy-looking question. Let A be the set of real x with x squared equal to one, and B, the real x with x to the power four equal to one. The curved E means, belongs to, the bold R means real numbers, and the colon means, such that. So, is A equal to B? If x to the power four is one, then x squared is one, or minus one. But can a real square be negative? Never! So x squared equals minus one is impossible. So x squared is one, and both sets are just minus one and one. Over the reals, A equals B, absolutely! But now change the universe. In the picture, the reals sit inside the complex numbers, the bold C on screen. There, x squared equals minus one is possible: x is i, or minus i, whose squares are minus one. So B becomes one, minus one, i and minus i, and the slashed equals sign says, B is not equal to A anymore. Same question, different universe, different answer! With me?

3:09Intervals: watch the endpoints

Now intervals, where the endpoints decide everything. Set A: x squared minus seven x plus ten, less than or equal to zero. It factors as x minus two, times x minus five, so x runs from two to five, both ends included. Closed interval, square brackets: the solid dots in the picture. Set B: mod of x minus three, the two vertical bars, is less than one. So x is between two and four, but the sign is strict, so both ends are excluded. Open interval, round brackets: the hollow circles. Clear? Now A minus B: what is left of A, once B is removed. Many students write four to five, and stop. But what about the number two itself? Yes, two survives! Two is in A, square bracket, but not in B, round bracket. Four survives too, for the same reason. So A minus B is the point two, union the closed interval four to five. The cup sign is union, okay? And B minus A? Every point of B is already in A, so it is the empty set, the circle with a slash. So the symmetric difference, the little triangle, is again two, union four to five. Endpoints, endpoints! Never forget them, alright?

4:33Topper's habit: check the data

Now let's take three sets. Every student in a class plays at least one of three games; call them A, B and C, okay? Forty play exactly one game, twenty play exactly two, and five play all three. What is the class strength? Simple, right? Look at the picture: the parts with no overlap, the parts where exactly two circles meet, and the centre. Every student is in exactly one of these. So the strength is forty plus twenty plus five: sixty five. But the question also says n of A, plus n of B, plus n of C, is ninety. Toppers never ignore extra data. Does it fit? See, this sum counts each student once for every game they play. So the forty count once, the twenty twice, and the five three times. Forty plus forty plus fifteen: ninety five, not ninety! So the data is inconsistent. If the question had given ninety five instead, everything would fit, and the strength would be sixty five. Always check the data, okay?

5:38Example 1: ordered pairs of subsets with fixed intersection

Now, example one. Find the number of ordered pairs, A comma B, where A and B are subsets of one, two, three, four, five, and A intersection B is exactly the set containing one. The sideways U with a line under it means, subset of, and the upside down U is intersection. Pause the video, and try it first. Done? Okay, let's check. Think element by element. The element one must be in A intersection B, so it sits in both sets. Only one way. Now the element two. Where can it go? In A, in B, in both, or in neither: four choices, right? No, no, careful! If two is in both, it enters A intersection B, and the intersection is no longer just one. So each of two, three, four and five has only three choices: neither set, A only, or B only, okay? The choices are independent, so we multiply. One, times three to the power four: eighty one ordered pairs. Nice and clean.

6:48Example 2: find n by double counting

Now a harder one. Thirty sets, A one to A thirty, each with five elements. And n sets, B one to B n, each with three elements. Both families have the same union, S: that is the big union sign. Every element of S lies in exactly ten of the A sets, and exactly nine of the B sets. Find the value of n, okay? Pause and try. Hint: count the same thing in two ways. Done? Okay, let's check. Remember the games? n of A plus n of B plus n of C counted each student once per game. Same idea here: we count pairs, an element and a set containing it. This is double counting. First, count the pairs set by set. Thirty sets, five elements in each. How many pairs? Thirty times five, one hundred and fifty, exactly! Now count element by element. Each element of S is in ten of the A sets, so it makes ten pairs: ten times n of S in all, and the dot means times. So ten times n of S is one hundred and fifty, and S has fifteen elements. Now the B sets: n sets, three elements each, so three times n pairs. Element by element: fifteen elements, each in nine B sets, so nine times fifteen, one hundred and thirty five. So three n is one hundred and thirty five, and n is forty five. Two counts, one answer. See?

8:26Quick revision: distinct elements, endpoints, double count

So, quick revision. One: count only the distinct elements, and check the universe and the endpoints. Two: think element by element, and multiply the choices. Three: count one thing in two ways, and always check the data. That completes our chapter on sets! Now go and try the practice problems on this page. You will find them easy, I promise. All the best!