Mixed Problems: Set 1 (Numbers, Words, Arrangements)

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. Take the six letters of the word school, and write every arrangement of them in dictionary order, just like the words in a dictionary. There are three hundred and sixty such arrangements. At what position does school itself appear? Will you write all three hundred and sixty to find out? No way, right? Today we solve a mixed set of problems on numbers, words and arrangements, and by the end you will find that position in a few lines. Questions like these come up often in JEE Main, and each one should take you only a minute or two. So, let's begin.

0:39Move 1: numbers from digits

So, move one, for numbers made from digits. Some places come with conditions, and the others are free. The rule is simple: fill the restricted places first, then the free ones. And remember, the first digit is never zero; otherwise the number loses a digit. Okay? Here is a problem from this set. We want the five digit numbers, with all digits different, that are divisible by five. So no repetition, right? Divisible by five means the last digit is zero or five. So two places are restricted: the last one and the first one. With me? First, say the last digit is zero. How many numbers do we get now? Zero is used up at the end, so the first digit can be anything from one to nine: nine choices. Then the three middle places take eight, seven and six choices; the dots mean times. Nine times eight times seven times six: three thousand and twenty four. Clear? Now the last digit is five. Same story, so three thousand and twenty four again, right? No, no, careful! This is the trap. Zero is still unused now, and it must not go first. So the first digit cannot be zero, and cannot be five: only eight choices. Then eight, seven and six for the middle places. Eight times eight times seven times six: two thousand six hundred and eighty eight. Got it? So the total is three thousand and twenty four, plus two thousand six hundred and eighty eight: five thousand seven hundred and twelve. That is why we split into two cases on the last digit. Okay?

2:21Move 2: words with repeated letters

Now, move two, for words. Arrange n letters, all used each time, where p of them are alike of one kind, and q are alike of another kind. The count is n factorial over p factorial times q factorial; the exclamation mark is the factorial sign. We divide because swapping two identical letters gives the same word. Right? Now, some questions say that certain letters must stay together. What do we do then? Yes, exactly! Glue them into one block. Treat the block as a single letter, and arrange it with the rest. Clear? And if they must not be together? Here is the topper's trick: do not count that directly. Count all the arrangements, and subtract the ones where they are together. Not together equals total, minus together. With me? Let's try it on the word balloon from this set, with the two O's not together. Seven letters, with two L's alike and two O's alike: seven factorial over two factorial times two factorial, that is one thousand two hundred and sixty. Now glue the two O's: six objects, with the two L's still alike, so six factorial over two factorial, three hundred and sixty. Subtract, and we get nine hundred. Perfect!

3:39Example 1: 4-digit numbers divisible by 4

Okay, example one, an easy one. How many four digit numbers, formed from the digits one, two, three, four and five, with repetition allowed, are divisible by four? Pause the video, and try it first. Done? Okay, let's check. First, when is a number divisible by four? Many students think: it just has to be even, so the last digit should be two or four, and that's it, right? No, no, careful! Even is not enough. Thirty four is even, but it is not a multiple of four. Okay? Here is the real test: a number is divisible by four exactly when its last two digits, taken together, form a multiple of four. So, by move one, we fill the last two places first. With me? So, from our five digits, the endings that work are twelve, twenty four, thirty two, forty four and fifty two. Forty four is fine, because repetition is allowed. Five endings in all. Clear? Now the first two places are free: any of the five digits, each time. Five times five: twenty five. Got it? So, five endings times twenty five: one hundred and twenty five numbers. Restricted places first, and the question is done in two lines!

5:02Example 2: rank of SCHOOL in dictionary order

Now, remember our puzzle from the start? Let's finish it. Find the rank of the word school among the arrangements of its letters in dictionary order. Every arrangement uses all six letters, and the two O's are alike, okay? Pause the video and try it. Done? Okay, let's check. First, what rank means. Rank equals one plus the number of words that come before ours in the list. So we only count the words before it. Right? Next, write the letters in alphabetical order: C, H, L, two O's, and then S at the end. Our word starts with S, the biggest letter. So every word that starts with a smaller letter comes before it. With me? Now, put C in front. How many words start with C? Five letters are left: H, L, O, O and S, with the two O's alike. So five factorial over two factorial: one hundred and twenty over two, sixty. It is the same for H and for L in front. So three times sixty: one hundred and eighty words. Okay? Next, O in front. Again five factorial over two factorial, sixty, right? Oh no, careful! One O is now used up in front. The five letters left, C, H, L, O and S, are all different. So it is just five factorial: one hundred and twenty. Divide only for letters that are still alike in what is left. Clear? Now S goes in front. After S, our word has C, then H, and no smaller letter is left for those places. Then the letters L, O, O are left, so the first such word is S, C, H, L, O, O, then comes S, C, H, O, L, O, and then our word, school. Got it? So the words before ours, inside the brackets: one hundred and eighty, plus one hundred and twenty, plus those two. That is three hundred and two words. Plus one for school itself. Rank three hundred and three! Puzzle solved, without writing three hundred and sixty words. Very good!

7:29Quick revision: digits, repeated letters and word rank

So, quick revision. Numbers: fill the restricted places first, and zero never goes in front. Words: n factorial over p factorial times q factorial; glue the letters that must be together, and for not together, take total minus together. Rank: one plus the number of words before it, counted place by place. Now go and try the three practice problems on this page. Give each one a minute or two, and you will find them easy. All the best!