Forming Numbers

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take the digits zero, one, two, three, four and five, with no repetition. How many five digit numbers divisible by five can you make? My student answered in ten seconds. Divisible by five, so the last digit is zero or five: two choices. The other four places, from the five digits left: five P four, one hundred and twenty ways. Two times one hundred and twenty: two hundred and forty. Sounds perfect, right? Well, twenty four of those numbers are fakes. Which ones? Hold that thought; we will catch them today. Look, questions on forming numbers come up often in JEE, and this one small slip costs marks. So, let's begin.

0:49Three rules for forming numbers

So, numbers formed from given digits. These are the most common counting questions of all, and three simple rules handle them. But first, a quick question. Zero, one, two, three, written in that order: is that a four digit number? No, it is not! That is just one hundred and twenty three, a three digit number. So, rule one: the first digit cannot be zero. Okay? Keep this rule in your pocket; it is the key to our puzzle. Now, suppose the number must be even. Which place should you fill first? The last place! Yes, exactly. Even or not depends only on the last digit, so fix that place first, and then fill the rest. So, rule two: a condition on the last digits, like the ones in the brackets, even, or divisible by five, by four, or by twenty five, fixes those places first. Right? And rule three. For divisibility by three or by nine, the test is on the digit sum, and the sum does not care about the order of the digits. So first list the sets of digits whose sum is allowed, and then arrange each set.

1:57Divisibility tests used in counting

Clear? Now, rules two and three need the divisibility tests, so let's put them all on one screen. By two: the last digit is even. By five: the last digit is zero or five. And by ten: the last digit is zero. Only one place to fix. Easy, right? Now tell me: for divisibility by four, which digits do you check? The last two digits, correct! They must form a multiple of four. And for eight, go one step further: the last three digits must form a multiple of eight. Got it? By twenty five: the last two digits must be zero zero, twenty five, fifty or seventy five. Only these four endings, okay? By three, and in the brackets, by nine: the digit sum must be divisible by three, or by nine. This is the digit sum test of rule three. With me? And two more. By six: divisible by two and by three, both tests together. By eleven: the alternating digit sum, add, subtract, add, and so on, must be divisible by eleven. Write these down. That's it!

3:00The zero trap

Alright, now the trap that catches the most students. When zero is among the digits, you must count carefully. Why? Because zero is allowed everywhere except the first place, okay? Way one: split into two cases. Case one, the restricted place, say the last digit, holds the zero. Then the zero is used up, and the first place is free. Case two, it does not. Then the zero is still waiting, and the first place loses one choice. Right? And here is a topper's trick, way two. Count all the arrangements, ignoring rule one, and then subtract the ones with a leading zero. Remember, a number starting with zero is just a shorter number in disguise. So throw those out! Clear?

3:45Example 1: 4-digit numbers from digits 1 to 7, and even ones

Let's start with an easy one. How many four digit numbers can be formed from the digits one, two, up to seven, without repetition? The dots on screen mean all the digits in between. And how many of them are even? Pause the video, and try it first. Done? Okay, let's check. No zero here, so rule one is not a worry. Four places, seven different digits: seven choices for the first place, then six, then five, then four. That is seven P four: the small seven at the top left of the P, and the small four at its bottom right. The crosses mean times. Seven times six times five times four: eight hundred and forty. With me? Now the even ones. Suppose you fill from the left as usual: seven, then six, then five. Now the last place must be even. How many choices are left for it? It depends! If two, four or six was already used in front, fewer even digits are left. No, no, that road is messy. Rule two: fill the restricted place first. The last digit is two, four or six: three ways. Okay? Now six digits are left for the other three places. Six P three: six times five times four, one hundred and twenty. Right? So, the even numbers: three times one hundred and twenty, three hundred and sixty. That's it! Last place first, and the question becomes completely simple.

5:20Example 2: 5-digit multiples of 5 from digits 0 to 5

Now, remember our puzzle from the start? Let's settle it. How many five digit numbers divisible by five can be formed from zero, one, two, three, four, five, without repetition? Pause the video and try it. And watch the zero! Done? Okay, let's check. Divisible by five, so the last digit is zero or five. Rule two: fix the last place first. And zero is among the digits, so we split into two cases. Okay? Case one: the last digit is zero. Now, can any digit that is left take the first place? Yes, exactly! The zero is already sitting at the end, so it can never lead. The digits one to five fill the other four places: five P four, one hundred and twenty. Got it? Case two: the last digit is five. Five digits are left, so five choices for the first place, right? No, no, careful! Zero is still in the pool, and rule one says zero cannot lead. So the first digit comes from one, two, three, four: four ways. Clear? Now four digits are left, zero included, for the three middle places. Zero in the middle is no problem. Four P three: twenty four. So four times twenty four: ninety six. With me? Total: one hundred and twenty, plus ninety six. Two hundred and sixteen. And now, our puzzle! My student said two times one hundred and twenty: two hundred and forty. But in case two he also counted numbers like the one on screen: zero first, then three empty boxes, then five at the end. The boxes take three of one, two, three, four: four P three, twenty four fake numbers. Two hundred and forty minus twenty four: two hundred and sixteen. Fakes caught! And see, that is way two from the zero trap: count all, then subtract the ones with a leading zero.

7:18Quick revision: first digit not 0, last places first

So, quick revision. The first digit is never zero, and the restricted last places are filled first. For digit sum tests, list the allowed sets of digits, then arrange them. And when zero is among the digits, split into cases, or subtract those with a leading zero. Now go and try the practice problems on this page. Watch that zero, and you will find them easy. All the best!