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ParthKohli
 one year ago
Continuing what I left off yesterday.
ParthKohli
 one year ago
Continuing what I left off yesterday.

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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3@ganeshie8 urgent...

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3\[x_1 + x_2 + x_3 = 20\]where\[3 \le x_1 , x_2 , x_3 \]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3That reduces to the coefficient of \(x^{20}\) in\[(x^{3} + x^{2} + \cdots + x^{20})^3\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3How do I calculate the above?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3@dan815 use elementary techniques please

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.1When it comes to these, it really depends on what exactly you want to do with it. There's a few different ways I can play around. Be more precise

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3\[=\left( x^{3}\cdot\dfrac{1x^{30}}{1x}\right)^3\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3\[= x^{9}(1x^{30})^3(1x)^{3 }\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3\[x^{9} (1  x^{90}  3x^{30} + 3x^{60})(1x)^{3}\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3\[= (x^{9}  x^{81}3x^{21} + 3x^{51} )(1x)^{3}\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3now the coefficient of \(x^r\) in \((1  x)^{n}\) is\[\binom{n+r1}{r1}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1you could also simply use stars and bars

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[x_1 + x_2 + x_3 = 20\]where\[3 \le x_1 , x_2 , x_3 \] is same as solving \[(y_1+3) + (y_2+3) + (y_3+3) = 20\]where\[0 \le y_1 , y_2 , y_3 \]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3hmm, yeah. is it \(y_1 + 3\) or \(y_1  3\)?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\(y_i=x_i+3\) so yeah it should be \(y_i3\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3ok, that's nice.\[\binom{n1}{k1}\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3I really wasn't trying to solve the problem ... I just want to know the approach to calculate the coefficient of \(x^k\) in huge expressions like the above one.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2id just use stars and bars

dan815
 one year ago
Best ResponseYou've already chosen the best response.2but sure ill bite, how does it come out to be the coeff of x^k again

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3oh sorry, I missed all the threes.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3\[y_1 + y_2 + y_3 = 29\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3should the number of nonnegative solutions be\[\binom{291}{31}\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.2oh nvm i get whwy its the coeff of x^20 in that expansion kinda neat trick

dan815
 one year ago
Best ResponseYou've already chosen the best response.2you would still use stars and bars to solve for the coeff i guess now, since theres no real nice way of expanding that

dan815
 one year ago
Best ResponseYou've already chosen the best response.2the way to think about this problem is the same as x1+x2+x3=k x1,x2,x3 >= 0

dan815
 one year ago
Best ResponseYou've already chosen the best response.2do you know how to solve that problem

dan815
 one year ago
Best ResponseYou've already chosen the best response.2allowed 3 what u basicaaly did is just allow more 1s in each place

dan815
 one year ago
Best ResponseYou've already chosen the best response.2by introducing 3 you just shifted your number index

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3ik, ik. there are various tricks to find the coefficient.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2ok so watcha looking for now

dan815
 one year ago
Best ResponseYou've already chosen the best response.2ya x1+x2+x3=29 is right

dan815
 one year ago
Best ResponseYou've already chosen the best response.2for x1,x2,x3 non negative solutions

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3ok, so nonnegative solutionsdw:1444026999220:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3eh, that thing was just meant to illustrate starsandbars for x1 + x2 + x3 = 4

dan815
 one year ago
Best ResponseYou've already chosen the best response.2heres a very easy way to work it out

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3\[\binom{29+31}{31}\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.2suppose x1,x2,x3 = 20 but u allow no 0 x1,x2,x3>=1

dan815
 one year ago
Best ResponseYou've already chosen the best response.2what does this really mean?

dan815
 one year ago
Best ResponseYou've already chosen the best response.2that means we must start with 1 in x1 x2 x3

dan815
 one year ago
Best ResponseYou've already chosen the best response.21 +k1+ 1 + k2 +1 +k3 = 20 so your new zero beginning equation is now k1+k2+k3=17

dan815
 one year ago
Best ResponseYou've already chosen the best response.2now u see how we went from the 1 case to 0 case

dan815
 one year ago
Best ResponseYou've already chosen the best response.2same logic from 0 to 3 case

dan815
 one year ago
Best ResponseYou've already chosen the best response.2just number indices here they are just symbols

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3i know what's happening here jesus

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3ok so by stars and bars, it turns out to be 465 however http://www.wolframalpha.com/input/?i=%28x%5E%283%29+%2B+x%5E%282%29+%2B+...+%2B+x%5E%2820%29%29%5E3 says x^20's coefficient is 402

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3ohhh I got where I went wrong lol

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1\[\binom{29+31}{31} \] this should be fixed no?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3i need \((x^{3} + x^{2} + \cdots + x^{26})^3 \)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3yay 465 is right baby

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Yeah see it should be \[\binom{29+31}{29}\] ye?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3well both of them are equal so it doesn't even matter

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Ah I see, you're too good Mr. Kohli!

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3how many solutions to \(x+y+z+t=29\) when \(x\ge 1, y\ge 2, z\ge 3, t\ge0 \)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3again we transform in the following manner\[x=a+1\]\[y=b+2\]\[z=c+3\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3how many solutions to \(x_1 + x_2 + x_3 + 4x_4 = 20\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3coefficient of \(x^{20}\) in\[(1 + x + \cdots + x^{20})^3(1 + x^{4}+x^8+\cdots + x^{20}) \]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3looks like you're back

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3I'm looking at a sample problem where they removed \(1x^{21}\) because it doesn't contribute to \(x^{20}\).

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3so can you use that trick here? I haven't got the real feel for it.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3\[\frac{\left(1  x^{21}\right)^3 }{(1x)^{3}} \cdot \frac{1 x^{20}}{1x^{4} }\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3now can we remove \((1x^{21})^3\) because it doesn't contribute to \(x^{20}\)?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3\[(1x)^{3} \left(1 + x^4 +x^8 + x^{12}+x^{16}+x^{20}\right)\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3now the coefficient of \(x^r\) in \((1x)^{n}\) is\[\binom{n+r1}{r1}\]is that right?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3or is it the opposite... coefficient of \(x^n\) in \((1x)^{r}\) is \(\binom{n+r1}{r1}\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3\[\frac{1}{(1x)^s}=\sum_{k=0}^{\infty} \binom{s+k1}{s1}x^k \]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3\[ \frac{1}{(1x)^s}=\sum_{k=0}^{\infty} \binom{s+k1}{k }x^k \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[\dbinom{n}{r} = (1)^r\dbinom{n+r1}{r1}\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3i'm appearing for an allindia test tomorrow and i'm really nervous :( so many expectations! how do I get the best out of myself?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[(1x)^{s}=\sum_{k=0}^{\infty} (1)^k\binom{s+k1}{s1}x^k \] right ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3hmm, but that \((1)^k\) shouldn't be there because there's another \((1)^k\) attached to \(x^k\) so they cancel out in a way
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