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I'm having a lot of trouble coming up with the situation/question for a combinatorial proof that asks to prove $$2^0 + 2^1 + \dotsb + 2^{n−1} = 2^n − 1$$

Using list counting. The question I came up with for the right hand side is: what is the number of lists length $n$ you can form with $2$ inputs, excluding the list where all the inputs are $2$.

vonbrand
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darrenz123
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  • "...you can form with 2 inputs, ...all the inputs are 2" (implying that $2$ is one of the possible inputs) You will find that it is more common to begin numbering things from zero rather than begin numbering things from one. If you have two inputs, they would be $0$ and $1$ rather than $1$ and $2$. That is not to say that counting from $1$ in scenarios like this is wrong per se, just that it is less common. Keep that in mind as you read answers. – JMoravitz Feb 18 '20 at 21:24

2 Answers2

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Hint: Think about binary strings of length $n$ with at least one $1$ ... & where the first $1$ occurs in the string.

Donald Splutterwit
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Every list of $n$ $0$'s and $1$'s has $i$ leading $0$'s where $0 \leq i \leq n$. There is $1$ list with $n$ leading $0$'s. And in general there are $2^k$ lists with $n-k$ leading $0$'s. The total number of lists of $n$ $0$'s and $1$'s is $2^n$.

Hans Hüttel
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