## 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)

1. anonymous

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

that is equal to 1771..

3. anonymous

@IrishBoy123 you know combinatorics?

4. IrishBoy123

looks right choosing with replacement, right?

5. anonymous

Yea I think so, I dont really understand the difference yet.. @IrishBoy123

6. IrishBoy123

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