Pnc

[errant tangle]

From an n sided polygon , an m sides polygon is to be constructed such that both the polygons don't have any common sides. Find no of ways to do this given that both the polygons are concave

[frosty swift]

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[errant tangle]

My thought process is making m vertices V1 V2 ...Vm and spaces between them on the n sided polygon be d1 d2 d3 ..dm

Now d1+d2+...dm= n-m
Where d1 can't be zero cuz then sides will be consecutive
Now we just have to find possible solns of the eqn for d1 d2 .. dm as natural nos. For which beggar coin theorem can be used

[normal river]

This is the same as picking m non adjacent vertices from n vertices.

[normal river]

Yess. That also works.

[normal river]

Here no d_i can be zero so as to not make adjacent, which meaningfully is the same thing.

[errant tangle]

Yeah I've specified that

[normal river]

Yeah then your method works.

[errant tangle]

Ok so ig 2025 sided polygon mein se hexagon banane ho with no side common so its 2024c5 right ?

[normal river]

a1+a2+a3+...a6=2025-6 Now a1=d1+1 and so on toh d1+d2+...d6=2013 ke solutions meaning 2018C5 aayega shayad

[errant tangle]

A1=d1+1 means ??

Oh right 2025-6-1c6-1

[normal river]

Matlab normal formula mein zero wale solutions bhi hote hain toh a1=d1+1 likh ke woh sirf a>=1 wale solutions nikalega

The n-r+1Cr-1 wala formula

[errant tangle]

Yeahh

Can u reply my dms pls

[errant tangle]

@normal river 2018c5 isn't the complete ans

It's only partially correct

Wanna give a try or should I share soln?

[normal river]

Ah I missed the concavity condition.

[normal river]

Note that for concavity point of concavity of n sided polygon must be chosen.

[errant tangle]

yeah

also theres repetition

[normal river]

Yeah precisely.

[normal river]

Division by 5! also.

[errant tangle]

naah

the complete ans is 2025 * 2018c5 * 1/6

[normal river]

Idt it should be concave but convex.

[normal river]

This is your wala method.

[errant tangle]

yeah i got confused b/w the two

[errant tangle]

generalizing it for m sided polygon from n sided polygon with 0 common sides we get
n*(n-m-1cm-1) 1/m right ?

idts we can generalize it for k common sides

[normal river]

You can

[errant tangle]

Can u tell hyw

How

I did hexagon out of a 2025 side polygon with 4 sides common and damn it was hard to process that shit

[normal river]

Assume that 1,2,3,4 and 5 are already chosen now do the d1+d2+... wala thing and multiply by 2025.

Idhar overcounting ka case nahi hai also ig.

[normal river]

@errant tangle Yo man this clear?

[errant tangle]

not really

in case of 4 common sides the 4 sides maybe all consecutive , 3 consecutive and 1 separated , 2 consecutive and other 2 consecutive

so i dont think generalising is possible/doable

[normal river]

Ah you're right this absolutely wrong.

[normal river]

Still ig you can generalize the casework.

Will try this later.

[errant tangle]

if there k common sides then no of possible cases will be ..?

[normal river]

Number of ways k can be written as a sum of integers.

Ah then thats not in jee tho

[normal river]

You're absolutely right.

[normal river]

But for 4 this will work.

[errant tangle]

we can do this by things included in syllabus but mushkil hoga
for k total cases will be
k consectuive sides
x1+x2 = k ke solns ,
x1+x2+x3=k ke solns ,
x1+x2+..... = k ke solns

ka sum

which in itself is hard to process ig

[normal river]

Yeah. It is actually something called partitions iirc.

[normal river]

Isse the thing is you get ordered pairs not unordered ones.

And there is no simple way to avoid repetition.

[errant tangle]

yeahhh

i have another pnc doubt , shoul i post it here or dms ?

[normal river]

...Anywhere ig?

[errant tangle]

+solved @normal river