Overview

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. Five friends sit in five chairs in a row. One of them refuses the first chair, and another refuses the last one. In how many ways can they sit? Many students take all the arrangements, subtract the two bad groups, and get seventy two. Sounds right. Well, it is wrong, and today you will see exactly why. This is the unit problem bank, where problems from the whole unit come mixed together, much like a real exam paper. So first, let me show you how to use it well. Let's go.

0:34How this chapter works

So, first, how this chapter is built. Part A, at JEE Main level: twenty seven solved problems, all mixed. Okay? Now, a question for you. When should you open Part A? Right after Chapter one, while the topic is fresh? No! Only after Chapters one to six, the whole unit. The little dash means one to six. And attempt it as a forty minute test, with a clock running. Right? Next comes Part B: longer problems, at JEE Advanced level. We will look inside it in a minute, okay? Then Part C, the concept checks. These are the statements tested most often in true or false form, or in which is correct form. Got it? And at the very end, a unit test of fourteen problems, answers only, no worked solutions. There, you check yourself honestly. Clear?

1:28Topper's trick: budget your 40 minutes

Now, a trick toppers use. Every problem carries a badge: easy, medium or hard. Count them first. Part A has seventeen easy, nine medium, and just one hard problem. With me? So, twenty seven problems in forty minutes. On average, how many minutes does each problem get? Forty divided by twenty seven: about one point five. That is the division sign, and the wavy equals sign means approximately, okay? So, roughly one and a half minutes each. Tight! And that is only an average. The easy ones must go really fast, so that the medium and hard ones get the extra time they need. Right?

2:11Part B: JEE Advanced level

Alright, now Part B, at JEE Advanced level. Twelve longer problems: two easy, seven medium and three hard. Give each one real time, okay? Look at what is inside. First, restrictions handled by inclusion and exclusion: count all the cases, then subtract the bad ones carefully. Our puzzle from the start uses exactly this idea. Right? Second, bounded distributions: objects go into boxes, but each box has an upper limit, at most so many in it. Our second example today is one of these. Got it? And then lattice paths, the shortest routes along a grid, and recurrences, where the answer for n is built from smaller cases. Clear?

2:55Part C: is the statement exactly right?

Now, one item from Part C, okay? Count the non-negative solutions of x sub one, plus dot dot dot, plus x sub k, equals n. The small letters below the x are subscripts, and the dots mean the terms in between. The claim: the answer is n C, k minus one, with the small n at the top left. The question mark asks: true or false? Oh no, careful! It is false. By stars and bars, the answer is n plus k minus one, C, k minus one. Students remember the k minus one at the bottom, and forget to add it at the top. One small slip, and the answer is wrong. That is what Part C catches. Keep this formula ready; our second example needs it. With me?

3:40Example 1: 5 in a row, A not first, B not last

Now, our puzzle from the start. It is a Part A problem, marked easy. Five people, A, dot dot dot, E, so A to E, sit in five chairs in a row. In how many ways, if A is not in the first chair, and B is not in the last? The people are different, so order matters. Start your clock, pause the video, and try it. Done? Okay, let's check. First, all the arrangements, with no condition. Five people in five chairs: five factorial, the five with the exclamation mark. One hundred and twenty. In the diagram, the big box U holds all of them. Okay? Now the bad cases. Let P be the arrangements with A in the first chair, shown by the circle P on the left. Fix A there, and the other four sit in four factorial ways: twenty four. So n of P, the number of arrangements in P, is twenty four. Right? In the same way, let Q be the arrangements with B in the last chair, the circle Q on the right. Again four factorial, so n of Q is twenty four. Got it? So the answer is one hundred and twenty, minus twenty four, minus twenty four. Seventy two, right? Hmm, look at the middle of the diagram, where the two circles overlap. No, no, careful! That is the trap from our puzzle. Seatings with A first and B last sit in both circles: P intersection Q, where the cap-shaped symbol means intersection. Fix A and B, and the other three sit in three factorial ways, that is six. We subtracted these six twice, once with P, once again with Q, okay? So, add them back once. One hundred and twenty, minus twenty four, minus twenty four, plus six: seventy eight. In the diagram: eighteen with only A first, eighteen with only B last, six with both, and seventy eight outside both circles. So, seventy eight, not seventy two. Puzzle solved!

5:39Example 2: 12 identical balls, 3 boxes, at most 6 each

Clear? Now a medium one from Part B, a bounded distribution. In how many ways can twelve identical balls be put into three different boxes, with no box holding more than six? Note: the balls are identical, the boxes are different, and nothing stops a box from staying empty. Pause the video, and try it. Done? Okay, let's check. An upper limit is hard to count directly. So, what do we count first? All the ways, with no limit at all, yes, exactly! Then remove the bad ones, just like Example one. Let x sub one, x sub two and x sub three be the balls in the three boxes. They add to twelve, and each x sub i is greater than or equal to zero, that sign with the line below. Stars and bars: twelve plus three minus one is fourteen, so fourteen C two: ninety one. Okay? Now the bad cases. Say box one holds seven or more. Simple trick: put seven balls into box one first. Five balls are left, for three boxes, with no limit. Five plus three minus one, that is seven, C two: twenty one. Right? And any one of the three boxes can be the bad one. So, three times twenty one: sixty three. The cross is the times sign, okay? Now, remember Example one? We had to add back the overlap. So here, do we add back the cases where two boxes both hold seven or more? Look: seven plus seven is fourteen, and the greater than sign says fourteen is more than twelve. We only have twelve balls! So the overlap is empty, and there is nothing to add back. Always check this before you add anything back. Clear? So, the answer is ninety one minus sixty three: twenty eight. Count everything, remove the bad cases, check the overlap. Same idea as Example one. Got it?

7:35Quick revision: the 40-minute test and the bad cases

So, quick revision. Part A: a forty minute test, after Chapters one to six. Part B: count all, subtract the bad cases, check the overlap. Part C: is the statement exactly right? Then, the unit test. Now open Part A, start your clock, and begin. After that, try the unit test at the end of the chapter. All the best!