Counting: Number of Elements in Unions

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. In a class of fifty students, thirty like cricket and twenty five like football. Thirty plus twenty five is fifty five. Fifty five students, in a class of fifty? Hmm, something is wrong here, right? Today we will find out exactly where those extra students come from. And look, this is not a small topic. In JEE, almost every word problem on sets is built on this one idea. So, let's begin.

0:32Two sets: add, then subtract the overlap

So, here is a set A, with all its elements shaded in blue. Now I bring in set B, and I want the total. Easy, isn't it? Just add n of A and n of B, right? Hmm, wait. Look at the middle. What happens to the elements that sit in both sets? Yes, exactly! They got counted twice, once inside A, and once again inside B, double counting! So what do we do? Simple. We subtract the overlap once. n of A union B equals n of A, plus n of B, minus n of A intersection B, okay? Write this down. This one line is the heart of today's lesson. Clear? And if the two sets have nothing in common? Then there is no overlap to subtract, so n of A union B is just n of A plus n of B, nothing more. Nice and simple. But only when the sets are disjoint, okay?

1:31More counts from the same numbers

Alright, now these same numbers give us much more. First, only A: elements in A, but not in B, okay? So take n of A, and throw out the overlap. n of A minus B equals n of A, minus n of A intersection B, that's it! Next, exactly one of the two sets. We call this A symmetric difference B: the little triangle you see on screen, okay? Now tell me, how many times should the overlap count here? Zero times, yes! But in n of A plus n of B, the overlap is counted twice, right? So, subtract n of A intersection B two times. Look, this is a classic JEE trap: students subtract it only once, and that gives at least one, not exactly one. Don't fall for it! Now the complement. The complement of A, written A prime, means everything in the universal set U that is not in A, okay? So, n of A prime equals n of U, minus n of A, simple! And now, the examiners' favourite: neither A nor B, nothing from either set. Where do these elements sit? Outside both circles, exactly! That means outside A union B, the shaded part. So, n of A prime intersection B prime equals n of U, minus n of A union B, okay? This is De Morgan's law at work. Remember this one; it comes again and again.

3:02Three sets: inclusion and exclusion

Okay, now three sets. Same idea, one level up. First, add n of A, n of B and n of C, all three. But now think. The centre region, which lies in all three sets: how many times did we count it? Three times! Absolutely. And every region in exactly two sets is counted twice. So, we subtract the three pairwise intersections. Now the regions in exactly two sets are fixed. But the centre? It was added three times and subtracted three times. So now it is counted zero times! Oh no, we lost it. So, just add the centre back once: plus n of A intersection B intersection C, okay? Now every region is counted exactly once. This is the famous principle of inclusion and exclusion. And the most common mistake in the exam? Forgetting this last plus term. Don't do that! Got it?

4:03Topper's trick: fill from the centre outwards

Now let me share a trick that toppers use. When a question gives you three sets, where should you start filling the Venn diagram? From the centre, always, and then outwards. Step one: the centre, g, the number of elements in all three sets. And if the question does not give you the centre, call it x, fill everything in terms of x, and solve for it at the end. Okay? Step two: the regions in exactly two sets. Region d is in A and B, but not in C, okay? So, d equals n of A intersection B, minus g. Why minus g? Because the centre also sits inside A intersection B, right? And the same way for e and f. Step three: the regions in only one set. For only A, take n of A and subtract d, f and g. These are the parts of A you have already filled, remember? And step four: the outside region is n of U, minus everything inside. Four lines, and the whole question is done. As simple as that.

5:11Example 1: 50 students, how many like neither sport

Now, remember our puzzle from the start? Let's finish it. In a class of fifty students, thirty like cricket, twenty five like football, and ten like both. How many like neither? Pause the video, and try it first. Done? Okay, let's check. Let C be the cricket fans and F the football fans. How many like at least one game? Thirty plus twenty five, minus the ten we counted twice. Forty five. See, mystery solved: those ten students were the extra ones in our fifty five! In the diagram: thirty minus ten, that is twenty, like only cricket; twenty five minus ten, that is fifteen, like only football; and ten like both. So, neither is fifty minus forty five. Five students. And if you did fifty minus fifty five and got a negative answer? That is the trap, right? Always subtract the overlap first.

6:10Example 2: three newspapers, A only and none of them

Now a slightly harder one, JEE level. Read it with me. In a survey of one thousand families, forty percent buy newspaper A, twenty percent buy B, and ten percent buy C, okay? Then, five percent buy A and B, three percent buy B and C, four percent buy A and C, and two percent buy all three. How many families buy only A, and how many buy none of the three? Pause and try. Hint: work in percentages, and change to families only at the end. Alright, only A first. Start with forty. Remove the five percent who also buy B, and the four percent who also buy C, okay? So thirty one percent, right? No, no, careful! This is the JEE trap. Both these groups include the two percent who buy all three, so those families got removed twice. Add them back once. Forty minus five minus four plus two: thirty three percent. That is three hundred and thirty families. Now, at least one newspaper. Which formula? Yes, inclusion and exclusion again! The singles add to seventy, the pairs add to twelve, and the triple is two. Seventy minus twelve plus two: sixty percent. So, none of the three is a hundred minus sixty: forty percent. Four hundred families. See how fast that was?

7:40Quick revision: union formulas for two and three sets

So, quick revision. Two sets: add both, subtract the overlap once. Three sets: add the singles, subtract the pairs, add the triple. And for word problems, fill the Venn diagram from the centre outwards. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!