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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

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  1. anonymous
    • one year ago
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    \[\binom{29+5-1}{5-1}\]

  2. anonymous
    • one year ago
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    we are discussing permutation and combination to solve these problems

  3. anonymous
    • one year ago
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    I came up with this C (33)(4) but what do I do with it?"

  4. anonymous
    • one year ago
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    Right, \(\dbinom {33}4={}_{33}C_4\), I'm using the binomial coefficient notation. Recall that \[{}_{n}C_k=\frac{n!}{k!(n-k)!}\]

  5. anonymous
    • one year ago
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    the thing is that I have the correct numbers but I do not know how to apply it.

  6. anonymous
    • one year ago
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    All you need to do is compute. Are you maybe confused as to how we arrive at \({}_{33}C_4\)?

  7. anonymous
    • one year ago
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    I am, how do you compute 33!??? 33x32x31x30?

  8. anonymous
    • one year ago
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    \[\frac{33!}{4!(33-4)!}=\frac{33!}{4!29!}=\frac{33\times32\times31\times30}{4\times3\times2\times1}=11\times4\times31\times30=\cdots\]

  9. anonymous
    • one year ago
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    ok 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

  10. anonymous
    • one year ago
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    Thank you so much!! how did you get the 11 x 4 x 31 x 30 ? that's the correct answer 40,920

  11. anonymous
    • one year ago
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    \[\frac{33\times32\times31\times30}{4\times3\times2\times1}=\frac{(\cancel3\times11)\times(\color{red}{\cancel8}\times4)\times31\times30}{\color{red}{\cancel4}\times\cancel3\times\color{red}{\cancel2}\times1}=11\times4\times31\times30\]

  12. anonymous
    • one year ago
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    thank you so much. I would have never got the right answer without your help.

  13. anonymous
    • one year ago
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    thank you so much. I would have never got the right answer without your help.

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