Solving Combinations Problem: 7 0s & 3 1s

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How many sequences are there consisting of seven 0's and three 1's? Example: 0101010000
I was thinking that I could solve this with combinations. However, I am unsure of how to use it in this case. If I remember correctly, the combinations are given by:

\begin{pmatrix}n \\ r\end{pmatrix}=\frac{n!}{r!\,\left(n-r\right)!}

Hmm. Any thoughts on this? Am I on the right track with the above?

Thanks.
 
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There are 2^{10} ways of selecting the digits and _{10}C_3 ways of having three 1's.
 
Tide said:
There are 2^{10} ways of selecting the digits and _{10}C_3 ways of having three 1's.
Alright. I was thinking that it didn't matter if they were the same number; I could still use the formula above. Thanks for the confirmation :smile:.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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