 Hello my dear students welcome to Centrum Academy YouTube channel where we bring for you every day a new problem in the out-of-problem Solving especially beneficial for the students who are preparing for J main and J advanced exams So what have we got for you in today's out-of-problem solving? So in today's question, we basically are gain dealing with a permutation and combination question So the question reads like this how many distinct parallelograms can be formed in an equilateral triangle of side length n? filled with n-square equilateral triangles of side length 1 each One such parallelogram is shown in the figure below in blue color So guys and girls as you can see very clearly in your diagram that there is an equilateral triangle Okay, so the big triangle that you see is an equilateral triangle of side length n and it has been filled up with Equilateral triangles of side length 1 each now Why does the question set us say that there are n-square equilateral triangles see guys and girls very simple If you see in row number one, there is one equilateral triangle in row number two There are three equilateral triangles in row number three There will be five equilateral triangles in row number four There will be seven equilateral triangles and so on and so forth So when you add these numbers One plus three plus five plus seven. Let's say for n terms Basically, what are you finding? You're finding the sum of an arithmetic progression whose first term is one common difference is two And you're summing this arithmetic progression or so as to say arithmetic series for n terms So the answer for this will be n by two to a Plus n minus one into the common difference which clearly boils down to n by two Two plus two n minus two so two and two gets cancelled and you have n by two times two n Even this two and this two will stand cancelled giving you n squared That's why the question set us says that there are n-square Equilateral triangles that are filled within this bigger equilateral triangle. Okay now moving on That's not my question actually my question is to count the number of distinct palerograms that can be formed in this entire structure All right guys So now what I want you to understand is the question can easily be solved by understanding few things number one Any parallelogram that you choose within this structure will fall under three categories So what are these categories number one? The parallelogram will have sides Parallel to any of the two sides of this bigger equilateral triangle for example The blue parallelogram that you see within this structure has got its sides parallel to AB So as you can see this side is parallel to AB and this side is parallel to AC Correct so likewise we can say another category could be a parallelogram whose sides are parallel to AB and BC Right so you can have a parallelogram whose sides is parallel to AB and BC Something of this nature something of this nature Correct so here you can see the sides are parallel to AB and BC and the third category is where the side is parallel to AC and BC Something of this nature Right so you can see here this parallelogram which I have shown with a highlighter is basically having its sides parallel to BC and AC all right now my viewers must appreciate that the total count of the number of Palergrams that will fall under each of these three categories would be the same. Yes, my dear students The number of parallelograms which will fall under case one Will be same as the number of palliograms which will fall under case two will be the same as that which will fall under case three so the Palliograms that you will be counting under case one and the Palergicam that you'll be counting on a case to and the Palliograms that it will be counting under case three They will all be equal right so it is sufficient for us to just count one of the cases because if we count one of the cases we can just multiply it with three and get our total answer alright so let us target let us target the case one first because if we target one of the cases the other case are automatically addressed okay now this problem can very easily be solved if you realize one more thing my dear students let us say I decide to extend this side length okay so let's say I decide to extend the side length AB I decide to also extend the side length AC okay alright so let me write a C here because I raise it okay and let me add one more row to this let's add one more row to this alright and now what I'm going to do here is I'm going to extend my mesh to add one more set of triangles down there yes so as you can see I'm extending the mesh to add one more set of triangles down there there you go my friends now this last row if you see here that last row will have got n plus two such points over here okay as you can see this will be my n plus 1th row and this last row will have n plus two points over here right so there will be a total of n plus two points in the last row so these counts will be having there will be n plus two points there now guys and girls I want you to understand one simple thing if I take a case of such a parallelogram which is having side parallel to AB and AC and let us say I extend this side length which I'm showing with a green color so let's say if I extend this side length you realize that on extension of that side length it will come to this point okay and let's say I extend this side length of the parallelogram as you can see on the figure it will meet over here so as you can see one of the points is here one of the point is here okay similarly if I extend this side of the parallelogram okay so that point will actually meet on this point on the last row and similarly if I extend this forward that will meet over here so basically what I see is that the parallelogram corresponding to the one which I had shown earlier is corresponding to four points chosen from those n plus two points down in the last line which you see in the figure right so these four points these four points the one which I'm putting a stick mark on the one which I'm putting a tick mark on they basically are the ones which correspond to that given parallelogram and my dear friends every parallelogram that you are going to pick up whose side is going to be parallel to AC and AB will have four points corresponding to it on the last line that you see in the figure so what does it mean it means that the number of parallelograms that you're going to get from here is going to be n plus two C four yes my dear friends so out of these n plus two points if you choose any four points it will correspond to one parallelogram which is basically having its sides parallel to AB and AC right so may I say all of these will correspond to the same number so this will also be n plus two C4 and this will also be n plus two C4 right so the total number of parallelograms the total number of parallelograms that you will end up getting my dear students is going to be three times n plus two C4 so this problem is solved so it was a question which was slightly tricky but with proper logic and reasoning you can easily crack such a problem so this is the answer to this question thank you so much for watching be safe and be healthy