Selection followed by Arrangement

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. We have seven consonants and four vowels, all different. We want words with three consonants and two vowels, and no letter repeated. Easy, right? Choose three consonants: thirty five ways. Choose two vowels: six ways. Thirty five times six, two hundred and ten words. Hmm. The true answer is one hundred and twenty times as many! So where did all those words go? Today we find out. And look, selection followed by arrangement comes up often in JEE problems on words and numbers. So, let's begin.

0:41Two stages: select, then arrange

So, first, one question. When we form a word, or a number, does the order of the letters matter? Of course it does! Swap two letters, and you get a different word. But the question has two stages hiding inside it. Stage one: which objects go in? That is a selection, and there the order does not matter. Choosing r objects from n different objects: n C r, the capital C, with n at its top left and r at its bottom right. With me? Stage two: arrange. The r chosen objects are all different, so they go in a row in r factorial ways. The exclamation mark, as you know, is factorial. Okay? Now tell me, do we add these two counts, or multiply them? Multiply, yes, exactly! Every single selection can be arranged in r factorial ways. So each selection gives r factorial arrangements, and we multiply. Right? And with no condition at all, n C r times r factorial is just n P r: the capital P, with n at its top left and r at its bottom right. The small dot on screen means times. Our old friend from permutations. So why bother with two stages? Because of conditions. Clear?

1:51Conditions: select group by group

Now, the conditions. Questions rarely say, choose any five letters. They fix how many of each kind must go in, for example three consonants and two vowels, like our puzzle. Okay? So we select from two groups: consonants from the consonants, and vowels from the vowels. Do we add these choices, or multiply them? Multiply, correct! Every choice of consonants can go with every choice of vowels. So select from each group, and multiply the counts. Got it? Then stage two. Now forget the groups. All the chosen letters are different, so we arrange all of them together, in one row. Right? And here is the trap from our puzzle. Students select perfectly, multiply perfectly, and then stop. They forget to arrange! That is exactly how we got only two hundred and ten. Careful, okay?

2:43Stage two depends on the rule

Now one more thing. Stage two is not always r factorial. When r different objects go in a row, in any order, then yes, it is r factorial. Got it? But now, from n people, choose a committee of four, with a leader. Stage one picks the four people. So what is stage two? Just four ways! We are not arranging the four people in a row. We only pick the leader from those four. So, n C four for the committee, times four for the leader. With me? And sometimes the rule allows only one order. Then stage two is just one way. Keep this in mind. It is the key to our second example. Clear?

3:20Example 1: words with 3 consonants and 2 vowels

Now, remember our puzzle from the start? Let's finish it. How many words of three consonants and two vowels can be formed from seven consonants and four vowels, with no letter repeated? Pause the video, and try it first. Done? Okay, let's check. Stage one, select. Three consonants from seven: seven C three, thirty five. Two vowels from four: four C two, six. Two groups, so multiply: two hundred and ten selections. Right? So the answer is two hundred and ten, isn't it? No, no, wait! That is the trap from our puzzle. Each selection is just five letters sitting in a group. They are not a word yet. Five different letters in a row: five factorial, one hundred and twenty ways. Okay? So, two hundred and ten, times one hundred and twenty: twenty five thousand two hundred words. See, mystery solved! That is why the true answer was one hundred and twenty times as many. Got it?

4:20Example 2: 4-digit numbers, increasing or decreasing digits

Now a slightly harder one. How many four digit numbers have strictly increasing digits? And how many have strictly decreasing digits? Pause and try. Hint: how many orders are allowed? Done? Okay, let's check. Strictly increasing, so no digit can repeat. Stage one: choose four different digits. Now stage two. Once you have your four digits, in how many ways can you write them in increasing order? Just one way! Yes, exactly! Smallest first, then the next one, and so on. There is no choice left. And the same for decreasing order. So stage two is one, and the count is just the selection. With me? So, ten digits, zero to nine. Choose any four: ten C four, two hundred and ten numbers with increasing digits, right? Oh no, careful! Look at zero. If zero is one of the four digits, in increasing order it would come first, and a four digit number cannot start with zero. So zero cannot occur. We choose from the other nine digits: nine C four, one hundred and twenty six. Clear? Now strictly decreasing. Can zero occur now? Yes, it can! In decreasing order, zero always comes last, never first. So all ten digits are allowed: ten C four, two hundred and ten. See? Two hundred and ten was wrong for increasing, but right for decreasing. Got it?

5:45Quick revision: select first, then arrange

So, quick revision. Select first, multiplying over the groups, and then arrange. r different objects in a row give r factorial; one fixed order gives just one way. And never stop at the selection, and watch the zero in numbers. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!