Overview

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Five different balls, three boxes, and no box may be empty. In how many ways can we put the balls in? One student says one hundred and fifty. Another student says twenty five. So, who is right? Here is the surprise: both of them can be right! Because I hid one small detail from you. Today you will see which detail, and the four questions that decide the method in every distribution problem. Questions like this come up often in JEE papers. Let's begin.

0:37Four questions before any formula

So, every distribution problem is the same picture: some objects go into some boxes. And before you touch any formula, you ask four questions. Question one: are the objects different, or identical? Okay? Question two: are the boxes different, or identical? And look at the brackets. Here a box can be a group, a person, or a room. Whatever receives the objects is a box. Right? Question three: may a box be empty? Or must every box get at least one object? With me? And question four: is there a limit on a box, some largest number it may hold? Four questions. Change any one answer, and the method changes. Clear?

1:23Question 1: different or identical objects?

Now, question one. Suppose the balls are identical and the boxes are different. Then what is the only thing that matters? Yes, exactly! Only how many balls go into each box. Which ball goes where means nothing. Okay? So let x one be the number in box one, x two the number in box two, and so on. The small numbers at the bottom are subscripts; they just label the boxes. So x i is the number in box i. Then x one plus x two, and so on up to x k, equals n, the total. Here k is the number of boxes. The three dots mean the pattern continues. Each way of filling the boxes is one solution of this equation. Got it? Now make the balls different, and keep the boxes different. What does each ball do now? Correct! Each ball makes its own choice: it picks a box. With k boxes, each object has k choices, so n objects give k to the power n: the small n at the top right. Right?

2:26Question 2: different or identical boxes?

Now question two. And this is the detail I hid in our puzzle! If the boxes are identical, you cannot tell one box from another. So with different objects, only the groups themselves matter: which balls sit together. Okay? Five different balls in three identical boxes, none empty: twenty five ways. You will count these in the section on identical boxes. So the second student was right, for identical boxes. With me? Now give the boxes names: box one, box two, box three. Take any one split of the balls into three groups. In how many ways can those groups go into the named boxes? Three factorial! The exclamation mark is factorial, as you know. No box is empty, so the three groups are all different, and we hand them out to the three named boxes in three factorial ways. Clear? So, twenty five times three factorial, that is twenty five times six: one hundred and fifty. The cross means times. Mystery solved! Same balls, but name the boxes, and the answer becomes six times as many. So always ask question two. Got it? And one warning. People, or teams called A, B and C, have names. So they are different boxes, never identical ones. Careful, okay?

3:43Questions 3 and 4: empty boxes and limits

Now question three: may a box be empty? For identical objects, no box empty means each x i is at least one, instead of at least zero. The sign on screen is greater than or equal to. Got it? For different objects into three different boxes, with no box empty, the count is three to the power n, minus three times two to the power n, plus three. The small dot means times. You will see where every term comes from in Example two. Right? And question four: is there a limit on a box? With identical objects, count all the solutions first, then subtract the ones that break the limit. With me? With different objects, say people into rooms, first list the box sizes that are allowed, then count each case. Four questions, four answers, and the method is fixed. Clear?

4:37Example 1: 10 identical balls into 4 different boxes

Now let's use our four questions. Example one is from the section on identical objects into different boxes. In how many ways can ten identical balls be put into four different boxes? And in how many ways with no box empty? Answer the four questions first. Then pause the video, and try it. Done? Okay, let's check. Question one: the balls are identical. Question two: the boxes are different. No limit on a box. So only the numbers in the boxes matter: x one plus x two plus x three plus x four equals ten. Right? First part: empty boxes are allowed, so each x i is at least zero. Then the count is n plus k minus one, C, k minus one: the capital C with numbers at its top left and bottom right, just like n C r. Here n is ten and k is four, so thirteen C three. Thirteen times twelve times eleven, over six: two hundred and eighty six. Okay? Now the second part: no box empty. Same formula, thirteen C three again, right? No, no, careful! Question three has changed, so the method changes. Now each x i is at least one, and the count is n minus one, C, k minus one: nine C three. Nine times eight times seven, over six: eighty four. See? From two hundred and eighty six down to eighty four, just from one question. Got it?

6:12Example 2: 5 different balls into 3 boxes, none empty

Now a harder one, from the section on different objects into different boxes. And look, it is our puzzle! In how many ways can five different balls be put into three different boxes? And in how many ways with no box empty? Pause the video and try it. Done? Okay, let's check. The balls are different, the boxes are different, and in the first part there is no limit at all. So each ball chooses one of three boxes: three to the power five, two hundred and forty three. Okay? Now, no box empty. Some of these two hundred and forty three ways leave a box empty. How do we throw them out? Very good! Choose the box to leave empty, in three ways, and put all five balls into the other two boxes: two to the power five ways each. Three times thirty two: ninety six. Clear? So the answer is two hundred and forty three minus ninety six: one hundred and forty seven, right? No, no, wait! This is the trap. Take the way with all five balls in box one. It leaves two boxes empty, so it was removed twice: once with box two empty, and once with box three empty. But it should go only once. There are three such ways, one for each box, so we add back three. With me? So, no box empty: two hundred and forty three minus ninety six plus three: one hundred and fifty. And that is the one hundred and fifty from our puzzle, found now by a completely different route. Perfect!

7:46Quick revision: the four questions before any formula

So, quick revision. Before any formula, ask: are the objects different or identical, and are the boxes different or identical? Then: may a box be empty, and is there a limit on a box? Change one answer, and the method changes. Now go through the sections of this chapter, and try their practice problems with these four questions in mind. All the best!