Hi everyone. So today I thought I'd go through a couple questions from India,
and these are from the 2018 JEE Advanced, Paper 1 mathematics section. So I
thought I might put these in my senior AMC playlist because they're sort of a
similar kind of difficulty. So pause the video here;
see if you can figure them out. Ok, let's start with question 17. So
the idea with these probability questions is that you want to have the
number of ways that none of the remaining students gets the seat previously
allotted to him, divided by the total number of ways of seating the students. So
that gives you the probability. Now we might start with working out the total
number of ways of seating the students. So not necessarily fulfilling that
condition. So that would just be 5 factorial. Now if you're new to combinatorics,
and you don't know what on earth I'm talking about, you would get that by
thinking of it like there's 5 choices for who goes in the first seat, and then
there's 4 remaining students to choose from for who goes in the next
seat, and then there's 3 to choose from for the next seat, and then 2 choices
and then 1 choices - 1 choice. So we call that five factorial with an
exclamation mark after it. So you can work out what that would be; that makes 120. So we're
gonna have like some number over 120, where this number is going to be the
number of ways that the remaining students can be seated. Now that's
actually what we would call a "derangement". So there is actually like a
formula for working that out like there's various ways you can calculate
it, but I'm not going to use that; I'm just gonna do it using systematic
listing, which just means, you know, I want to write out all the different ways of
doing it, but I want to do it in a particular order, so that I can be sure
that I haven't skipped any. So I would probably start with... let's go
like seat 1, 2, 3, 4, 5. So student 1 has to sit in seat 1. Now
we've got to sort of work out all the ways that these remaining students can be
seated. So I can't put student 2 here because
they're not allowed to have the seat previously allocated to them, so I'm
gonna start by putting student 3 in seat 2. And then I'm gonna put student 2 here.
And then I can't put 4 here, so I'm gonna put 5 there. And then 4's gonna go
there. Now I've done that in that very specific way because I want to list them
sort of in increasing numerical order. So I want to start with like the lowest
possible number that I can put here and then the lowest possible number I can
put here and so on. So that's gonna be the first way to do it. Now let's think
of the next way. So I can put 3 here again. Now I can't put this in a different
order, like I can't have 4 5 because then they'd be in the same seats as
before, so that's the only way to do it going 1 3 2, so now I have to think about
changing this number, so 4 would be the next lowest number that I could put
there. And now we've got 5 and 2... I can't do 2 5 because then the 5 is
gonna be in the wrong column, so we have to do 5 2, and then that's the only way to
arrange them with 1 3 4 at the start. So now let's try... the next one up from there
would be 1 3 5, and that leaves 2 and 4 which have to go in that order because
the 4 can't go there. Okay and then there's nothing else that I could change
this to; there's no higher number, so that's actually all the ways of doing it
with 1 3 at the start. Now we can actually kind of do a shortcut here
because... so there's gonna be like all the ways of doing it with 1 4 at the
start, and then all the ways of doing it with 1 5 at the start,
and because of a sort of a symmetry of the situation,
I actually know that there's going to be as many here as there are here,
and as many here as there are here. So I could actually just do a shortcut here
and say that there are 9, because there's gonna be 3 there, 3 there,
and 3 there. But um I'll actually list them all out for you, in case that was
too much of a stretch of the imagination. So what's the first one we could put
here? We could put 2 there and then that leaves 3 and 5, which we have
to put in that order. So the next one... I can't put 3 here; it'll have to be
5 there, and then we've got 2 and 3 left, so they could go in that
order, or they could go in that order. And then
that's all of the ones that have 1 4 at the start. And then we can go 1 5. I
could put 2 there, and then 3 4. Now we can... we can't have 3 there but
we can have number 4 there, and then I can have 2 3 in that order, or
3 2. And then that's everything; that's all of them. And you can see I've listed
them in increasing numerical order, like if this was the number thirteen thousand
two hundred and fifty four, and so on - think of them like that - then they're in
increasing order. And that's sort of essential to systematic listing because,
if you don't write them out in order, if you just sort of start listing them
randomly, then you can never be sure that you've gotten every
single one. But now we can be sure, because I've listed them in order, so I
know like there was nothing that could go between these two because of the way
that I've very... I've listed them in a very organised way. So that's the idea
with these kinds of questions. Okay so we've got 9 different ways, so it's
just... that's 9. Nine ways out of 120. So we could simplify that fraction down
because both of those are divisible by 3, so we can make it 3 over 40,
so that would be A. All right let's do question 18. So
first challenge in this one is just trying to work out what the question is
asking. And for some of you who may not be so familiar with set notation,
that might be particularly challenging. So I'm just going to delve into what on
earth they're talking about here. So Ti is the event that students Si and Si+1
do not sit adjacent. Let's think about what that means. So T1
for instance would mean that the students S1 and S2 don't sit next to
each other, and T2 is going to mean that students 2 and 3 don't sit next
to each other, and T3 is going to mean that students 3 and 4 don't sit next to
each other, and so on. Now this little symbol here is set
intersection, so you've probably seen, you know, Venn diagrams before, where you've
got set A and set B and then this bit in the middle is the intersection of the
two sets, so you would express that as A intersection B and that's as opposed to
A union B which means like the whole thing, the combination of the two sets.
Now if we're talking about events in probability then you can think of this
symbol as meaning "and". Like it's saying that T1 happens AND T2 happens AND T3
happens AND T4 happens. So if we're talking about probability and you see
this, think of that as "and"; that one would mean "or". Okay so what it's saying
is that students 1 and 2 don't sit next to each other, and students 2 and 3 don't sit
next to each other, and students 3 and 4 don't sit next to each other, and
students 4 and 5 don't sit next to each other. So it's asking for the probability
of that. So again I'm just going to do systematic listing; I'm just going to
write out all the different ways that that could happen. So like the previous
one it's gonna be something over 120, so we just got to work out the
number of ways that this can happen. So seats 1 2 3 4 5.
What we could do is we could put student 1 there and now I can't put student 2 here
because student 1 and 2 aren't allowed to sit next to each other, so I
could put student 3 here. Again I'm gonna be listing them from sort of smallest
numerical value to largest, so that I can be sure that I get every single one. So
what's the next one that could go here? I can't put 2 there because 2 can't sit next to
3, and also I can't put 4 there because 3 can't sit next to 4. So let's put number 5
there, and then that leaves 2 and 4, so we can put them in that order. I actually
can't put them the other way around because 5 can't sit next to 4. So that is
the only way to do it with 1 3 at the start because once we put 1 and 3 in there
we were sort of... we had to do 5 2 4; there was no other way to do that. So
next I'm going to do 1 4. So I could put 2 there, and then that leaves 3 and 5. I
can't put 3 there because 2 and 3 can't sit next to... next to each other,
so I'm going to put 5 3. And then that was the only thing I could put there
because 3 and 5 couldn't sit next to 4, so I'll have to change that. Let's go up to 5.
Now if I put 2 there, then the only choices I have for there are... like
there's 3 and 4 left, and 3 and 4 can't sit next to each other, so I
actually can't do 2 there. Let's try putting 3 there, and then that leaves
2 and 4; I can't do that either because neither of those can sit next to
the 3. So there actually... there's nothing I can put there, because I can't
put 4 there either, because 5 and 4 can't sit next to each other. So
that's actually all the ways we can do it with number 1 at the start. So I'll
just draw a little line there to separate them. So let's try putting 2
at the start. Now we can't have 1 there, we can't have 3 there,
I could put 4 there, and then we could have 1 there, and then 3 5, or I
could do 2 4 1 5 3. Or... what else can I do?
I can't put 5 or 3 there, because they can't be next to 4, so
that's all the ways of doing 2 4 something, but maybe we could try 2 5.
I could do 2 5... I can't do 2 5 1 because that
leaves 3 and 4 and they can't sit next to each other, but I could do 2
5 3, and then 1 4. And then that's it
for the ones that start with a 2. Now let's try 3 at the start, so I could
do 3 1... I can't do 2 there but maybe 4, and then that leaves 2 5.
Or I could do 3 1 5 2 4.
And then... now I can't do 2 or 4 there because they can't go next to the 3, but we
could do 5 and then 1... What's left? 2 and 4. I could have 4 2; I can't
do 2 4, because the 1 and the 2 can't go next to each other. Or we could have...
we can have 3 5 2 and then that leaves 1 and 4. I can't put the 1 there
but I can put the 4 there. Now what you might notice at this point is that these...
this one is what I would call a sort of a mirror image of this one.
By "a mirror image" I mean... well it's not really a mirror image,
but it's um... what you would get replacing 1 with 5, and 2 with 4, leave 3 as a 3, sort of
reversing the order of the numbers. Like if I took that one, replace the 1
with 5 and the 5 with 1 and so on,
I would get this one. And similarly with doing the "mirror image" of this one; that
would give you that one. Now we could do a similar thing with... each of these will
have a mirror image. So like this one for instance would turn into 4 1 3 5 2, and
you know, we know that that's gonna be another possible way, so each one of
these is gonna have a mirror image which is gonna be something that I haven't
listed before. So you could actually kind of stop when you reach this point. It's
gonna be these 7 ways and their mirror images. I'll just list them all
out in case, you know, you bothered to list all of them out and you want to
check against my ones. So this one's going to become
4 2 5 1 3. That one's gonna be 4 2 5 3 1. That one's
5 2 4 1 3. And that one's gonna be 5 3 1 4 2. So we
get these 14 ways. So you might have done it by listing each of those out
individually or you might have sort of stopped there and thought "Hey, I can
double this!" So either way we're gonna get 14 out of 120, and then we can
simplify that down, because both of those are even so we get 7 out of 60, so that
one is the answer. All right guys I hope you enjoyed this one. If there was
anything that you didn't understand that you'd like me to clarify, then just
leave a comment below. And if you'd like me to make more videos on JEE questions
then give this video a thumbs up and subscribe. And I'll put a link to where
you can download these papers in the description. And if there's any
particular question from the JEE or whatever other exam that you want me to
go over then just let me know in the comments.