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well that's just choosing 2 spaces out of 9.... cool i got it it's (9 chooses 2)... brilliant...
umm... take x_xxxxxx_xxx_xxx.... so there are groups (x) (xxxxxx) (xxx) (xxx).... let a=(x) b=(xxxxxx) c=(xxx) d=(xxx)..
the sum of these groups or variables is 13. and i took the c and d as the same as there is no constraint like a,b,c,d are distinct....
that's easy just take 13 x's in a line put 3 dashes in between them... it gives me 4 groups of x's... well sum of groups in any such arrangement will give me 13... these groups are my variables a,b,c,d.... right...?
the minimum value of any of the variables is 1 and maximum is 8 right.... so if i take x as say 1 then the rest of the two variables are to be selected from the remaining x's right?
yeah... i get it.... so these variables (groups)
sorry my bad... i was counting the dashes as well in my mind... anyways... the answer to this step is again 10... you mean that any such arrangement with two dashes has the sum of such groups equal to 10 right...
you mean just add the 3+4+3? or like the total number or arrangements....
if the first thing is what ur asking me to do then it's 12
they are greater then 3 but less then 4 in each group of x's @bobbym
@agnishom no i don't but i have seen a formula related to that problem.... i just can't remember it...
yup... two dashes in the arrangement...
go it.... but why 10... why not 9 or 11...
well any suggestions on how to solve this current problem....
if u do know any good book or site to practice combinatorics and probability it would be great if you could share them with me..
yes exactly i am a 12th grade student... pretty good at calculus... but i suffer in combinations... well the constraints on x,y,z are clear the N starts from 1 and not 0 (isn't the set containing 0 as well called 'whole numbers') anyways.... tell me the basic idea i might actually think of getting the solution even without the things u mentioned... (stars and dashes etc...)
yeah i understand a little about generating functions....
if x,y,z belong to N (set of natural numbers)
then the number of solutions to this equation x+y+z= 10 are.... well the answer is given as 36 (from 9C2.... combinations)
can anybody help me in finding the number of solutions.... also help me find the idea to find the number of solutions in any general equation
if an electric field E and magnetic field B exists in a frame of reference F1.... for another frame of reference F2 this magnetic and electric fields and B* and E* respectively where F2 is moving with a constant speed V with respect to F1.... what is the relationship between E, E* and B, B*
i am still stuck with it i don't see any relevance other then the sin {a+(n-1)d/2} that comes in the sum of sines i am not getting to the proof....
well i am stuck with my post on sum of sines with angles in A.P. i still didn't get to prove it can you help me on that?
well... is there a function which cannot be expressed in the form of polynomials...
well i need the Taylor series only for basic applications....
thanks bob... appreciate it i will work on the problem following your advice....
osculating polynomials? can you elaborate please
we have this series sin a + sin a+d + sin a+2d + sin a+3d+.......+sin a+(n-1)d = [sin (nd/2) sin{a+(n-1)d/2) ] over sin (d/2)... this is given in my book... but i don't trust them i want to find out the answer to this series on my own... and the angles need not necessarily add up to 180 degree.... bob can you give any advice on this? a quick one if you are in a hurry
i posted my problem on the forum please help me find the answer for it.... it's bugging me from a long time...
i need to find out the formula to find the sum of sines whose angles are in A.P. in as many ways as i can... can anybody help me?
P.S. i should be apply such techniques to find the sum of cosines or tangents whose angles are in A.P.
can i apply it to find the sum of sines where the angles are in Arithmetic progression? i mean i don't even know how to find the sum of such series by any method.... so the Taylor series can be helpful (if it can be used..)