Mixed Problems: Set 2 (Selections, Divisors, Geometry)

Transcript

0:00Introduction

Hello everyone! Okay, one small challenge before we start. How many divisors does eighteen hundred have? Let's try listing them: one, two, three, four, five, six, eight, nine, ten, twelve, and we are nowhere near the end! Slow, right? And so easy to miss one. By the end of this video, you will count them in a few seconds, and even say how many are even, and how many are perfect squares. This is set two of our mixed problems: selections, divisors and geometry. Questions like these come up often in JEE Main. So, let's begin.

0:42Tool 1: selections with at least or at most

So, this set needs three tools. Tool one: selections. When a question says at least, or at most, there are two routes. Add the cases one by one, or take the total, minus the cases you do not want. Pick whichever is shorter. Right? Now, one case often needs some men and some women in the same committee. Do we add their choices, or multiply them? Multiply, yes, exactly! Every choice of men can go with every choice of women. So within one case, we multiply. Clear? But two different cases can never happen together. So over the cases, we add. With me? In short: and means multiply, or means add. Keep this ready for example one. Okay?

1:28Tool 2: divisors from the prime factorisation

Tool two: divisors. Always start by writing the number N as a product of primes: p one to the power a one, times p two to the power a two, and so on, up to p k to the power a k. Here p one up to p k are different primes. The small one, two and k written low, on the p's and the a's, are subscripts, just labels; the raised a's are the powers; and the dots mean the pattern continues. Got it? Now, a divisor of N uses only these same primes, and never more of each one than N has. So, in a divisor, how many choices do we have for the power of p one? Yes, exactly! Zero, one, two, and so on, up to a one. That is a one, plus one choices. The same for every prime, so we multiply: a one plus one, times a two plus one, and so on, up to a k plus one. Power zero for every prime gives the divisor one, and it counts. Clear? Now two special kinds you will need in this set. An odd divisor has no factor two at all. So the power of two must be zero: just one choice. Right? And an even divisor needs at least one two. So the power of two starts from one, not from zero. One choice fewer. With me?

2:46Tool 3: geometry is just choosing points

Tool three: geometry. Take n points in a plane, with no three collinear, that is, no three on one straight line. Any two of them fix a line, and any three fix a triangle. Order does not matter, so we just choose. n C two lines, and n C three triangles: the capital C, with n at its top left, and two or three at its bottom right. Okay? Now, a quick one from this set. Eight points lie on a circle. How many triangles have their vertices among them? Eight C three, correct! Three points on a circle are never on one straight line, so every choice of three gives a real triangle. So eight C three: fifty six triangles. Got it? And quadrilaterals work the same way: choose any four of the points. Eight C four: seventy quadrilaterals. That's it!

3:40Example 1: committee of 5 with at least 3 men

Right? Alright, example one, marked easy. We choose a committee of five from six men and four women, with at least three men. In how many ways? Pause the video, and try it. Done? Okay, let's check. It is a committee, so order does not matter: we choose. Now, at least three men. So we take three men and two women, and we are done, right? No, no, careful! At least three means three, or four, or five men. Three cases, not one. Read the condition first, every time. Okay? Case one: three men from six, and two women from four. Six C three is twenty, and four C two is six. One case, two groups, so we multiply. The small dot means times. Twenty times six: one hundred and twenty. With me? Case two: four men and one woman. Six C four is fifteen, and four C one is four. Fifteen times four: sixty. Right? Case three: all five are men, and no woman. Six C five: six ways. Clear? Now add over the cases: one hundred and twenty, plus sixty, plus six. One hundred and eighty six ways! Multiply within a case, add over the cases: exactly tool one. Got it?

5:07Example 2: divisors of 1800, even and perfect squares

Now, remember our challenge from the start? Let's finish it. This one is marked medium. For eighteen hundred, find the number of divisors, the number of even divisors, and the number of divisors that are perfect squares. Pause the video, and try all three. Done? Okay, let's check. Tool two says: factorise first. Eighteen hundred is two cubed, times three squared, times five squared. The small raised numbers are the powers. Right? So the number of divisors: three plus one, times two plus one, times two plus one. Four times three times three: thirty six divisors! Challenge solved, in a few seconds, without listing a single one. With me? Now the even divisors. What can the power of two be now? One, two or three, yes! Not zero, because an even divisor needs at least one two. So three choices for the two, and still three and three for the others. Three times three times three: twenty seven even divisors. Okay? Last part: perfect squares. In a perfect square, every power is even. Two is cubed in eighteen hundred, so the power of two can only be two. Just one choice, right? Oh no, careful! Zero is even too. Two to the power zero is just one, and that is allowed. So the power of two is zero or two: two choices. The same for three, and for five. Two times two times two: eight square divisors.

6:42Quick revision: selections, divisors and geometry

Clear? So, quick revision. Selections: multiply within a case, and add over the cases. Divisors: factorise into primes, add one to each power, and multiply. And geometry: with no three points collinear, just choose them: n C two lines and n C three triangles. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!