On pg. 55 in Nielsen and Chuang, it's said that:
the $|0\rangle + |1\rangle$ product can be well approximated by a superposition of states of the form $|0\rangle ^{\otimes n(1−p)/2}|1\rangle ^{\otimes n(1−p)/2}$.
I'm confused about how we get the exponents for $|0\rangle$ and $|1\rangle$, namely, $\otimes n(1−p)/2$.
Forget about the whole $n(1-p)$ for a minute.
For simplicity, let $k$ be even and think of the product $\tfrac{(|0\rangle+|1\rangle)}{\sqrt{2}}^{\otimes k}$ like it's the binomial $(x+y)^k$ with commuting indeterminates $x$ and $y$.
When you expand the binomial $(x+y)^k$ the monomial $x^jy^{k-j}$ with the largest coefficient i.e. the monomials that appears the most frequent number of times in the expansion before collecting like terms are those with ${k/2}$ $x$'s and $k/2$ $y$'s. To be exact there are exactly $k\choose k/2$ such monomials.
So if someone was going to draw randomly picked monomials from $(x+y)^N$ with say $N$ really really big the most likely outcome is a monomial with ${N/2}$ $x$'s and ${N/2}$ $y$'s.
This is exactly the same thing that's going on with the product state. When you expand the state $\frac{(|0\rangle+|1\rangle)}{\sqrt{2}}^{\otimes k}$ the superposition with the highest probability is one with $N/2$ $|0\rangle$'s and $N/2$ $|1\rangle$'s, which after reordering we could write as the product state $|0\rangle^{\otimes k/2}|1\rangle^{\otimes k/2}$.
The same thing goes on when $k$ is odd since, in the grand scheme of things with say $k$ really big, it all comes out in the wash.
Now it should be clear what is going on pg.55 of N&C, by replacing $k$ with $n(1-p)$ for $0\leq p\leq 1$.
