Thank you very much for your detailed explanation for both of my doubts.

I understood your points above this message:

I understand that the different numbers in the Pascal's triangle show the values of C(n,r) where n and r depend upon the row and distance from the leftmost side of the triangle.

Is the Pascal's triangle's base on the right? I think that is the case as it follows the general trend: The number in the next row is the sum of numbers in the current row. I think, here, the trend is switched to columns. Or are we supposed to follow this format as opposed to the usual triangle?

"It is unclear whether your book says anything about C(n,k) for n < 0, but since you didn't mention it I assume it doesn't." - Even I'm not sure about the intention of the authors. It plainly gave this without any explanation: C(n,r)=0 if r ∉ {0,1,2,3,...,n}

I've learnt all the three parts of your definition earlier, but I never knew that we can use any integral value for n and r in C(n,r).

Regarding the second question I asked:

I understand why a bijection (one-one and onto) between two sets implies the two sets have the same number of elements. However, I haven't encountered cases in combinatorics questions where I need to prove a bijection exists from one set to the other. Or in other words, I'm unable to see how it's related to this question. If possible, could you give an explanation or suggest a resource where I could learn about this?

As my book just gave the formulas (n-1)! and (n-1)!/2 for the circular permutation without any explanation, I learnt about it from this answer on the main site.

That was how I understood why it's (n-1)! permutations when clockwise and anticlockwise arrangements are counted separately and (n-1)!/2 when both directions are considered as the same thing. I didn't know to prove that before reading that answer as I didn't even fully understand what circular permutations were.

But now, the way I'd arrive at (n-1)! and (n-1)!/2 is similar to the method described in that answer. That method was satisfying for me and so I proceeded. But I don't know whether it's rigorous enough.

I can see that you've given me some hints to arrive at the formula rigorously in this message, but I'd like to try it after clearing the doubts I mentioned in this message.

@user21820: Kindly reply when you find time. Thank you for your time and help :-)

Anyone please have a look at my question.

Consider point A(5, 2) and variable points B(a, a) and C(b, 0). If the perimeter of triangle ABC is minimum, then find a, b.

I know that for minimum length, path followed by light ray is to be traced but how to apply it here.

@GuruVishnu It's quite common for high-school level textbooks to be unclear about what they mean. But I personally think it is far better to be precise. Regarding Pascal's triangle, it doesn't matter whether you think of it upright (the usual) or sideways (in the way I presented it). What is important is simply that you realize the best way to extend it to negative n.

@GuruVishnu Sadly, that's the problem with high-school level textbooks. I haven't seen any deal properly with combinatorics. Let me explain the circular permutation one and you will know what I mean.

Let W be the set of ways of arranging n distinct beads on a circle (equally spaced), where we consider two ways the same iff one can be obtained from the other by rotating it around the circle.

Let P be the set of permutations of {1..n−1}, meaning the set of lists that each has each k∈{1..n−1} in exactly one position.

First number the n beads 1 to n. For each arrangement A, define f(A) to be the list obtained by reading clockwise from just after bead n until just before bead n. Then f(A) is a permutation of {1..n}. (Note that this f is well-defined because we consider arrangements to be the same under rotation.)

Now we need to prove that f is injective and surjects onto P. For any arrangements A,B such that f(A) = f(B), reading the beads clockwise from just after bead n gives the same for both A and B, so clearly A and B are the same (up to rotation). Thus f is injective. I'll let you check for yourself that f surjects onto P, since it is trivial.

If this explanation is completely satisfactory to you, you can proceed to try counting circular permutations with some same beads, where we only consider arrangements the same under rotation. Don't go to flipping yet, as you are not yet ready for the complications that arise.

But if you want a 100% rigorous mathematical formalization of the above, you would have to deal with equivalence classes as follows.

To make it a bit simpler, we can use {0..n−1} instead of {1..n}. Each permutation of {0..n−1} can be defined as a bijection from {0..n−1} to itself. Circular permutations can be defined as equivalence classes of permutations of {0..n−1} modulo the following equivalence relation: Given permutations A,B of {0..n−1}, define A ~ B iff ∃c∈Z ∀k∈{0..n−1} ( A(k) = B((k+c)%n) ). Here x%n is defined as the unique r∈{0..n−1} such that n | x−r. Now check that ~ is an equivalence relation.

Let P(n) be the set of permutations on {0..n−1}. Since ~ is an equivalence relation on P(n), we can construct P(n)/~ := { { A : A~B } : B∈P(n) }. This P(n)/~ is the mathematically precise formulation of "circular permutations of {0..n−1}". So what we want to count is the size of P(n)/~.

You can define the bijection equally precisely using the equivalence relation. For each X∈P(n)/~, define f(X)∈X such that f(X)(n−1) = n−1, which exists because X = { A : A~B } for some B∈P(n), and B(c) = n−1 for some c∈{0..n−1}, so we can define f(X) to be ( {0..n−1} k ↦ B((k+c−n+1)%n) ), so that f(X)~B and f(X)(n−1) = B(c), giving f(X)∈X and f(X)(n−1) = n−1 as desired. Finally, define g(X) to be f(X) restricted to {0..n−2}.

If you are a mathematics major, it would be instructive to prove as an exercise that this f bijects from P(n)/~ to P(n−1), to fully formalize the intuitive reasoning. I don't mean to say that you must do this for each combinatorics question, but you must learn how to do it if you wish to truly grasp the underlying mathematics.

However, if you are only at the high-school level, feel free to ignore whatever I said about equivalence classes.

@Doubtnut Given B, you can find the optimal C using the reflection trick. Best to reflect C to get C'. Then you want to find minimum total length of BC and BC'. Applying the reflection trick again gives you the optimal B.

@user21820, can you please elaborate more. I've to reflect C about which line and why I've to fing BC'.

@Doubtnut That's for you to try to figure out. I don't give out answers.

By the way, how did you write @Doubtnut in orange box.

@Doubtnut If you hover your mouse over a message, a right-arrow button appears. That is the reply button. Click on it then type your message, and it will automatically make a ping (with an orange box) plus a link to the replied message. When you ping me, I also see an orange box, but if you don't click the reply button then it will not link to the earlier message.

@user21820, I'm using mobile site.

@user21820 I thought the extended Pascal's triangle would be an inverted triangle. But I was surprised that it extends only on one side (towards left):

Is there any reason for this extended triangle to not be symmetrical about the vertical axis?

@GuruVishnu Exactly that is why we need to see Pascal's triangle via the recursive relation that I gave, rather than trying to view it upright.

When I was younger, I tried to extend the upright one upwards, and didn't arrive at the 'correct' version.

I thought negative values of n had no meaning. At least you tried it before someone asked you to do so :-)

So the table form of the Pascal's triangle you gave is known as 'Recursive relation', am I right?

Thank you for your explanation of circular permutations. I think I've to read it some more times to understand it completely. By the way, I've learnt some basics about equivalence classes in the chapter relations. I just know a relation is an equivalence class if it's transitive, reflexive and symmetric and nothing more than that. This seems to be a bit different for me.

@GuruVishnu Technically speaking, "recursive relation" is the relations that generate the table. In binomial coefficients, the recursive relation is:

And the base cases are covered by:

@GuruVishnu A relation is an equivalence relation iff it is reflexive and symmetric and transitive, as you stated.

@user21820 Ok. Thank you for letting me know that.

But I suspect your book did not even bother to teach equivalence classes. The definition of equivalence classes is exactly the one I used:

> Given any set S and equivalence relation ~ on S, the set of equivalence classes is S/~ := { { A : A~B } : B∈S }.

Notice that for any two objects A,B∈S, either A~B or ¬A~B.

If A~B, then { X : X~A } = { X : X~B }, which intuitively means that A and B have (i.e. are in) the same equivalence class.

If ¬A~B, then not only { X : X~A } ≠ { X : X~B }, but in fact { X : X~A } ⋂ { X : X~B } = ∅, which intuitively means that the equivalence classes for A and B are disjoint.

You can try proving these basic results to get a feel for the foundational structure of equivalence classes.

Note that { { A : A~B } : B∈S } is a set of equivalence classes, so each equivalence class is included only once. This is the precise method to formalize the notion of "consider ... the same iff ...".

@GuruVishnu: If you want to clarify any specific point in more detail, feel free to ask. I am not doing every tiny step so that you have something to try, but if you get stuck let me know.

Thank you for your time in explaining these. I'll go through these and ask further if I've any doubts.

You're welcome!

I'm now going to move on to chemistry. I'll come to this after some time and ping you if I face any issues. Sorry if I'm making you wait so long. It's just because of the fact, I find it easier to load inorganic chemistry into my mind during this time :-)

No problem!

Hey guys