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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)
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im just thinking this equation, \[\left(\begin{matrix}n+r1 \\ r\end{matrix}\right)\] 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that is equal to 1771..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@IrishBoy123 you know combinatorics?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0looks right choosing with replacement, right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yea I think so, I dont really understand the difference yet.. @IrishBoy123

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0https://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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