Selections with Conditions

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. There are twelve people, and we must choose a committee of five. Easy so far, right? But two of these twelve refuse to serve together. Now how many committees can we make? Hmm. Committees with the first one but not the second, with the second but not the first, with neither of them. It gets messy very fast. By the end of this lesson, you will do it in one line. And look, conditions like at least, at most and never together come up often in JEE problems. So, let's begin.

0:37At least, at most: two routes

So, first, the words at least and at most. Suppose a team must have at least one girl. Tell me, if you count it directly, how many cases is that? Many cases! One girl, two girls, three girls, and so on, right up to a full team of girls. That is the direct route: list the cases, and add them. Okay? Now the other route. First count all the selections, with no condition at all. That is the total. Then subtract the cases you do not want. We call this the complement. So, for at least one girl, which teams are not allowed? Only the teams with no girl at all. Yes, exactly! So, at least one equals the total, minus, in the brackets, none at all. Many cases on one side, just one subtraction on the other. See the difference? Clear? So here is the topper's habit. Before you start counting, look at both routes, and take whichever is shorter. Sometimes it is the direct route, sometimes it is the complement. We will see both in today's examples. Got it?

1:43Two particular objects never together

Now, back to our puzzle. Two particular people must never be together. That means neither of them is in, or exactly one of them is in: two cases. Now think about the opposite: both of them in. Which one is easier to count? The opposite, yes! Both in is just one case. But first, all the selections. Choosing r objects from n different objects, where order does not matter, is n C r. That is the capital C, with n at its top left and r at its bottom right. With me? Now, the selections with both of them in. Put the two inside first. They already fill two places, right? So we only choose the other r minus two, from the remaining n minus two. That is n minus two, C, r minus two. Okay? So, never together equals n C r, minus, n minus two, C, r minus two. The total, minus the selections we do not want. That's it! Clear? Now our puzzle: n is twelve, and r is five. Twelve C five is seven hundred and ninety two. With both of them in, choose the other three from the other ten: ten C three, that is one hundred and twenty. So, seven hundred and ninety two minus one hundred and twenty: six hundred and seventy two committees. One line, as promised! Right?

3:02Several groups: multiply, then add

Now, many questions mix groups: boys and girls, men and women. Say one case needs some girls and some boys in the same team. Do we add their choices, or multiply them? Multiply, correct! Every choice of girls can go with every choice of boys, right? So within one case, multiply the choices from each group. But different cases, like one girl or two girls, can never happen together. So over the different cases, we add. Okay? In short: and means multiply, or means add. Remember this one, and the long questions become really simple. Got it?

3:43Example 1: committee of 4 with at least one man and woman

Alright, example one. Read it with me. We choose a committee of four from five men and four women, with at least one man and at least one woman. In how many ways? Pause the video, and try it first. Done? Okay, let's check. Directly, we need one man and three women, or two and two, or three men and one woman: three cases. Now, which route is shorter here? The complement! First, the total. Nine people in all, all different, and order does not matter in a committee. Nine C four: one hundred and twenty six. Right? Now, which committees break the rule? Only two kinds. All men: five C four, that is five. And all women: four C four, that is one. Oh no, careful here! Many students subtract the all men committees and forget the all women one. Four women, four places: just one way, but it still counts! So, one hundred and twenty six, minus five, minus one: one hundred and twenty. Two subtractions instead of three cases. See how fast that was?

4:51Example 2: team of 5 with at least 2 and at most 2 girls

Now a slightly harder one. We choose a team of five from six boys and five girls. In how many ways, if it has, part A, at least two girls, and part B, at most two girls? Pause and try. Hint: for part A, count the opposite. Done? Okay, let's check. Part A first. Eleven different people in all, so the total is eleven C five: four hundred and sixty two. With me? Now, the opposite of at least two girls. That means no girls, so we subtract just the all boys teams, right? No, no, wait! Fewer than two girls means zero girls, or one girl. Zero girls: all five from the six boys, six C five, that is six. One girl: five C one, times six C four. The small dot on screen means times. One case, two groups, so we multiply: five times fifteen, seventy five. Okay? So, four hundred and sixty two, minus six, minus seventy five: three hundred and eighty one. Forgetting the one girl case is the classic trap here. Don't fall for it! Now part B: at most two girls. Direct route, or complement? Direct, and here is why. At most two means zero, one or two girls. And two of those cases we already found in part A: six, and seventy five. Right? Only the two girls case is new. Choose two girls from five, and three boys from six: five C two, times six C three. Ten times twenty: two hundred. Now add over the cases: six, plus seventy five, plus two hundred. Two hundred and eighty one. Multiply within a case, add over the cases: exactly what we learnt! Got it?

6:41Quick revision: at least, never together, several groups

So, quick revision. At least and at most: add the cases, or take the total minus the opposite, whichever is shorter. Two particular objects never together: n C r, minus, n minus two, C, r minus two. And with several groups: multiply within a case, add over the cases. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!