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dan815
 one year ago
just for fun, see if you think of some ways to prove
\[\sum_{k=0}^{m1}\binom{m1}{k}=2^{m1}\]
dan815
 one year ago
just for fun, see if you think of some ways to prove \[\sum_{k=0}^{m1}\binom{m1}{k}=2^{m1}\]

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A combinatoric interpretation of this fact is given by counting subsets of different sizes of a set \(S\) of \(m1\) elements. Since we count the number of subsets of size \(k\) for \(0 \le k \le m1\), this sum must be equal to the number of subsets of \(S\), which is \(2^{m1}\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Consider \(m1\) place for a subset. For each subset it can either include or not include an element. For each element, there are \(2\) possibilities. Multiplying these together we get \(2^{m1}\) subsets.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You can use binomial theorem.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0As well as power set interpretation

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Just for the sake of alternative, using pascal's rule can be enlightening too \[\binom{n}{k}=\binom{n1}{k1}+\binom{n1}{k} \] dw:1435740725845:dw Induction step goes like this \[\begin{align} \sum\binom{m}{k}&=\sum\left[\binom{m1}{k1}+\binom{m1}{k}\right]\\~\\ &=\sum\binom{m1}{k1}+\sum\binom{m1}{k}\\~\\&=2^{m1}+2^{m1}\\~\\&=2^{m} \end{align}\]

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.0\[A=\left\{ a_1,a_2,a_3,...,a_{m1} \right\}\\card(a)=m1\\ \left(\begin{matrix}m1 \\ 0\end{matrix}\right)+\left(\begin{matrix}m1 \\ 1\end{matrix}\right)+...+\left(\begin{matrix}m1 \\ m1\end{matrix}\right)=total \space \space subsets \space of A\\=2^{card(A)}=2^{m1} \]
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