Question 72
There are $2n$ light bulbs arranged in two rows, A and B, numbered from $1$ to $n$ in each row. Some of the bulbs, possibly none, are on and the rest are off; such a configuration is called a state. Two states are different if a bulb is on in one and off in the other. A state is called good if the same number of bulbs are on in row A as in row B.
Let $EB$ be the total number of good states and $ET$ the total number of states. Prove that
$$
\frac{EB}{ET}=\frac{3\cdot5\cdot7\cdots(2n-1)}{2^n n!}.
$$
Let $EB$ be the total number of good states and $ET$ the total number of states. Prove that
$$
\frac{EB}{ET}=\frac{3\cdot5\cdot7\cdots(2n-1)}{2^n n!}.
$$