I am completely lost on how to achieve this. I have no idea where to start, nor do I know what to use to find to prove this problem. Can someone help me with this?
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What is $C(n,i)$? – Feb 03 '16 at 16:05
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I assume binomial coefficient (i.e. "choose" function), in which case the equation holds. – parsiad Feb 03 '16 at 16:05
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1@sigmabe: alternative form of $\binom{n}{i}$ – robjohn Feb 03 '16 at 16:05
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It is a choose function, correct. – JCMcRae Feb 03 '16 at 16:08
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Algebraic proof:http://math.stackexchange.com/questions/1649712/prove-the-formula-sum-k-1n-k-binomnk-n-cdot-2n-1-for-all-integer – NoChance Feb 10 '16 at 22:29
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The left hand side counts the number of ways to select $i$ of $n$ available people to serve on a committee, then choose the chairperson for that committee for committees ranging in size from $1$ to $n$ people. The right hand side counts the number of ways of choosing a chairperson for a committee, then choosing a subset of the other $n - 1$ people to serve on the committee.
N. F. Taussig
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@J-Play The Binomial Theorem states that $$(a + b)^n = \sum_{k = 0}^n \binom{n}{k}a^{n - k}b^k$$ Setting $a = b = 1$ yields $$2^n = \sum_{k = 0}^{n} \binom{n}{k}$$ The term $\binom{n}{k}$ is the number of subsets of size $k$ of a set with $n$ elements. Hence, a set with $n$ elements has $2^n$ subsets. – N. F. Taussig Feb 03 '16 at 17:08
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On the left hand side, $\binom{n}{i}$ is the number of ways of selecting $i$ of the $n$ people to be on the committee and $i$ is the number of ways of selecting a chairperson from among the $i$ people in the committee. The summation counts the number of ways of selecting committees with a chairperson ranging in size from $1$ to $n$. On the right-hand side, there are $n$ ways of choosing a chairperson and $2^{n - 1}$ possible subsets of the remaining $n - 1$ people who could serve on the committee with the chairperson. – N. F. Taussig Feb 03 '16 at 18:00
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HINT
Consider that you are trying to find all possible permutations of $A,B,C$ of the length $n$ when $A$ can be used only once. The number of such permutations would be $n2^{n-1}$.
Full Solution
Then, consider you are trying to choose where $B$ and $A$ are first. Deciding where $B,A$ would be would be $C(n,a)$ if the sum of the number of $B$s and the number of $A$s equaled $a$. Then, deciding where $A$ would be among these $a$ numbers would be $a$. In conclusion, for each $a$ there are $a \times C(n,a)$ number of ways. Therefore, $n*2^{n-1}$ would be equal to the left hand side.
Chad Shin
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