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anonymous
 one year ago
How many solutions (using only nonnegative integers) are there to the following equation?
x1 + x2 + x3 + x4 + x5 = 29
anonymous
 one year ago
How many solutions (using only nonnegative integers) are there to the following equation? x1 + x2 + x3 + x4 + x5 = 29

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\binom{29+51}{51}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we are discussing permutation and combination to solve these problems

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I came up with this C (33)(4) but what do I do with it?"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Right, \(\dbinom {33}4={}_{33}C_4\), I'm using the binomial coefficient notation. Recall that \[{}_{n}C_k=\frac{n!}{k!(nk)!}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the thing is that I have the correct numbers but I do not know how to apply it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0All you need to do is compute. Are you maybe confused as to how we arrive at \({}_{33}C_4\)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am, how do you compute 33!??? 33x32x31x30?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{33!}{4!(334)!}=\frac{33!}{4!29!}=\frac{33\times32\times31\times30}{4\times3\times2\times1}=11\times4\times31\times30=\cdots\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok I have \[\frac{ 982,080 }{24(29!) }\] where do you get 11 x 4 x 31 x 30 this is so frustrating ok let me calculate 29! so it would be 29 x 28 x 27 x 26 = 570,024 so the answer would be ....i cant get it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much!! how did you get the 11 x 4 x 31 x 30 ? that's the correct answer 40,920

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{33\times32\times31\times30}{4\times3\times2\times1}=\frac{(\cancel3\times11)\times(\color{red}{\cancel8}\times4)\times31\times30}{\color{red}{\cancel4}\times\cancel3\times\color{red}{\cancel2}\times1}=11\times4\times31\times30\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you so much. I would have never got the right answer without your help.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you so much. I would have never got the right answer without your help.
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