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anonymous

  • one year ago

You have four large cointainers, the first one holds only 10 cent coins, the second 5 cent coins, the third 1 cent coins and the last one 0.5 cent coins. In how many ways can you select 20 coins from these?(combinatorics)

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  1. anonymous
    • one year ago
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    Im just thinking this equation, \[\left(\begin{matrix}n+r-1 \\ r\end{matrix}\right)\] cause im gonna make 20 selections(r = 20) among 4 elements(n = 4) \[\left(\begin{matrix}23 \\ 20\end{matrix}\right)\] is this completely wrong?

  2. anonymous
    • one year ago
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    that is equal to 1771..

  3. anonymous
    • one year ago
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    @IrishBoy123 you know combinatorics?

  4. IrishBoy123
    • one year ago
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    looks right choosing with replacement, right?

  5. anonymous
    • one year ago
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    Yea I think so, I dont really understand the difference yet.. @IrishBoy123

  6. IrishBoy123
    • one year ago
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    https://en.wikipedia.org/wiki/Combination it's not something i do/ have done very much; and as soon as you delve a little deeper i am no use. for this, the link explains it better than i could. ganeshie is great at this stuff [as he is at most things] and there are plenty of others here too that can help of course, i am always willing to help whenever i can :p

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