Question

probability(choosing) question:
if (5C0)+(5C1)+(5C2)+(5C3)+(5C4)+(5C3)=2^k,
find k.

Answer 1

\[If\left(\begin{matrix}5 \\ 0\end{matrix}\right)+\left(\begin{matrix}5 \\ 1\end{matrix}\right)+\left(\begin{matrix}5 \\ 2\end{matrix}\right)+\left(\begin{matrix}5 \\ 3\end{matrix}\right)+\left(\begin{matrix}5 \\ 4\end{matrix}\right)+\left(\begin{matrix}5 \\ 3\end{matrix}\right) = 2^{k}, find k\]

Answer 2

isn't the last term 5C5 ? otherwise you don't get an even number.
to solve, simplify the terms on the left to numbers, add them up, and expects a power of 2. For example 5C0= 1. continue for 5C1, etc

Answer 3

nope, in book is 5C3 (the last one) prolly miss print then i guess =/

Answer 4

\[\sum_{c=0}^{6}{\left(\begin{matrix}5 \\ c\end{matrix}\right)}=2^k\]
\[32=2^k\rightarrow 2^5=2^k\]
k=5

Answer 5

0_o

Answer 6

If last term isn't 5C5 then k=5.39232