Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Five friends, A, B, C, D and E, and five chairs in a row. Now, friend A refuses the first chair, and friend B refuses the last chair. All the arrangements: one hundred and twenty. With A in the first chair: twenty four. With B in the last chair: twenty four. One hundred and twenty minus forty eight is seventy two. Done, right? Hmm, not so fast. Seventy two is wrong, and six arrangements have gone missing. By the end of this class, you will know exactly which six. This is Part A of our unit problem bank, at JEE Main level, and restrictions like these come up often. So, let's begin.

0:48Part A: a 40-minute test

So, here is what Part A holds. Twenty seven solved problems, all at JEE Main level. Short, sharp questions, okay? And they are mixed: numbers, words, round tables, selections and distributions, all together. Nobody tells you which chapter a question comes from, just like the real exam. Right? The plan: finish Chapters one to six, then attempt Part A as one forty minute test. So today, two tools that several Part A problems need, then two of its problems, one easy and one harder. With me?

1:24Tool 1: none of the bad cases

Tool number one, from Chapter five: inclusion and exclusion. Capital N is the total: all the arrangements, with no condition at all. That is the whole box, marked U for the universal set. Inside it, two circles, P and Q: just names for two bad properties, the things we do not want. Okay? Now, how many arrangements are bad? Easy, just add the two circles, right? Hmm, look at the middle. What goes wrong there? Yes, exactly! The middle got counted twice: once inside P, and once again inside Q, double counting! So we subtract it once. The vertical bars mean, the number of arrangements in. The cup is union, and the cap is intersection. So, the number in P union Q equals the number in P, plus the number in Q, minus the number in P intersection Q, clear? But in Part A, we usually want the good ones: none of the bad properties. That is the shaded part, outside both circles. So take the total, and throw out the union. N, minus the number in P, minus the number in Q, plus the number in P intersection Q, that's it! Look at that last plus. Forget it, and the middle gets thrown out twice. That is exactly the trap in our puzzle. Got it? And here is the topper's trick for more properties. S one, with a small one at its foot, is the sum over the single properties; S two is the sum over pairs; S three, over triples. The number with none of the properties is N, minus S one, plus S two, minus S three, and so on; the dots mean the signs keep alternating. Right?

2:59Tool 2: round tables and blocks

Tool number two, from Chapter two: the round table. Look at the figure. In picture one, marked with a small roman one, four people, A, B, C and D, sit around a table. In picture two, everybody has moved one seat, and still the arrow says, same. Everybody moved! So why is it the same arrangement? Because at a round table, only the relative positions matter. Look, everybody still has the same neighbours on the left and on the right. So fix one person, and arrange the rest. For n different people, that gives n minus one, in brackets, factorial; the exclamation mark means factorial. That is the blue line under the pictures. And picture three, the mirror image? At a table, it is a different arrangement. Only for a necklace is it the same, and that is why the red line halves the count. At a table, never halve. Clear? Now, suppose some people must sit together. The block method: glue them into one block, and treat the block as one single unit. Okay? Then, inside the block, they can still swap places. So multiply by the orders inside the block: for a pair, two factorial, that is two. Forgetting this two is a common slip.

4:12Example 1: 5 in a row, A not first, B not last

With me? Now, remember our puzzle? Let's finish it. Five people, A to E, sit in five chairs in a row; the dots on screen stand for the letters in between. In how many ways, if A is not in the first chair, and B is not in the last? Pause the video, and try it first. Done? Okay, let's check. Five different people, one in each chair, no condition: five factorial, one hundred and twenty. That is capital N, the whole box. Clear? Bad property P: person A sits in chair one. Fix A there, and the other four fill the other four chairs: four factorial, twenty four. Okay? Bad property Q: person B sits in chair five, the last one. The same way, four factorial, twenty four. Right? Now, the puzzle answer was one hundred and twenty, minus twenty four, minus twenty four. But tell me, which arrangements did we throw out twice? The ones in the middle: A in the first chair and B in the last, both at once! Fix both of them, and the other three fill three chairs: three factorial, six. These six sit inside P and inside Q, so they were thrown out twice. Clear? So, add them back once. One hundred and twenty, minus twenty four, minus twenty four, plus six: seventy eight. There are our six missing arrangements! In the diagram: eighteen in only P, eighteen in only Q, six in both, and seventy eight in the good region outside. Perfect!

5:47Example 2: 8 at a round table, two pairs not neighbours

Now the harder one, where both tools work together. Eight people sit at a round table. Persons A and B must not be neighbours, and C and D must not be neighbours. In how many ways? Pause the video, and try it. Hint: use both tools. Done? Okay, let's check. Eight different people at a round table, no condition: eight minus one, factorial, that is seven factorial. Five thousand and forty. That is our total. Right? Now bad property P: A and B sit together. Glue them into one block. The block plus the other six people: seven units. So, seven factorial ways, right? No, no, wait! Seven units at a round table, not in a row. Tool two: seven minus one, factorial, that is six factorial, seven hundred and twenty. And inside the block, it can be A then B, or B then A: so times two. The dot on screen means times. Two times seven hundred and twenty: one thousand four hundred and forty. Okay? Bad property Q means C and D sit together. Exactly the same count: one thousand four hundred and forty. Clear? Now, tool one says: subtract both, then add back the overlap. So what is the overlap here? Both pairs together, at once! Two blocks, plus the four other people: six units. Six minus one, factorial: five factorial, one hundred and twenty. Then times two, and times two again, one for each block: four hundred and eighty. With me? So, none of the bad properties: five thousand and forty, minus both bad counts, two thousand eight hundred and eighty, plus four hundred and eighty. Two thousand six hundred and forty. Forget the plus four hundred and eighty, and you fall into the same trap as our puzzle. Don't! Got it?

7:46Quick revision: none of the bad cases, round tables, blocks

So, quick revision. One: none of the bad cases is N, minus the number in P, minus the number in Q, plus the number in P intersection Q; never forget that last plus. Two: at a round table, n minus one, factorial, and that holds for the units after gluing too. Three: for people together, glue them into a block, then multiply by the orders inside. Now go and attempt all twenty seven problems of Part A as one forty minute test. You will do well, I promise. All the best!