I dont know even how to start

@Sourav_Dey sir , @Jagdish_Singh sir , @Hari_Shankar sir , @Mayank_Chowdhary , @Raghudevram_Singh

# Doubt from Permutations and combination

**Tushar_2019**#1

Well what's the answer given ?

Also in what form is it asked ?

Coz I feel that the answer must be some multiples of a number like 19a and we are asked to find a

So we can use a bit approximations

Since Imo it is too difficult to find the actual answer

I just came to the conclusion that that the body diagonal of the cuboid will pass through 4 lattice points (so we can say that the answer is a multiple of 3)

So basically we need to find the solution for a cuboid of dimensions 50×108×125

It's a nice question it seems

**Tushar_2019**#7

@Viraam_Rao , @Azimuddin_Sheikh , @pratyaksh_tyagi , @Chirag_Hegde , @Rafe_Moin1 see this!

**Sourav_Dey**#8

150+324+375 - GCD(150, 324) - GCD(150, 375) - GCD(324, 375) + GCD(150, 324, 375) = 768(ans )

**Tushar_2019**#14

Consider the set of all triangles OAB where O is the origin and A and B are distinct points in the plane with non negative integer coordinates (x,y) such that 41x + y = 2009. Find the number of such distinct triangles whose area is a positive integer.

@Sourav_Dey sir , @Mayank_Chowdhary , @Raghudevram_Singh , @Viraam_Rao

**Mayank_2019_1**#15

Now that we need to find the difference to be a even number

So we can say that

Even-even =even

Odd-odd = even

**Mayank_2019_1**#16

And there are 24×24 ways of getting both x1 and X2 as even number

And 25×25 ways of getting x1 and X2 as odd number

Also that x1 and X2 belong to (0,49]

Also we need to divide the whole by 2 as we will have repeating cases

**Tushar_2019**#18

Since 2009 is not even, ((x_1)-(x_2)) must be even, thus the two x's must be of the same parity. Also as u said maximum value for x is 49 and the minimum is 0. There are 25 even and 25 odd numbers available for use as coordinates and thus there are (_{25}C_2)+(_{25}C_2)=600

\therefore 600 \over 2 = 300 should be our required answer but answer is 600

@Mayank_Chowdhary

**Mayank_2019_1**#19

if you select 2 out of 25 we won't get repeating cases

I actually did something silly ... Leave what I did for now

You are correct we need to select 2 numbers out of the given 25 numbers so it's 25C2 so why would be need to divide by 2 again

See if we think like this :

If we first select any one out of 25 so we have got 25 choices and for selecting second one we have 24 choices so total 25×24

But here there are cases that are repeated.

Eg. In the first chance In the first round I choose let's say 24 and in the second round 22 and now in the next chance I select 22 in the first round and 24 in the second round.

Hence we need to exactly divide by 2

**Tushar_2019**#20

There are 19 flags of which 10 are identical blue flags, and 9 are identical green flags and there are two distinguishible flagpoles. Find the number of distinguishable arrangements using all of the flags in which each flagpole has at least one flag and no two green flags on either pole are adjacent

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@Mayank_Chowdhary , @Sourav_Dey sir @Ayan_Agrawal @Viraam_Rao @Raghudevram_Singh

general formula is not working here ??