= Solution
For the <deep-water gravity wave>, the <phase velocity> and <group velocity> follow from its <dispersion relation>:
$$
v_p=\frac{\omega}{k}=\sqrt{\frac gk},\qquad v_g=\frac{d\omega}{dk}=\frac12\sqrt{\frac gk}=\frac{v_p}{2}.
$$
The <wave packet> travels at $v_g$, while the individual crests travel at $v_p$. Let its effective length be $D\simeq5\lambda$, where $\lambda=2\pi/k$ is the carrier <wavelength>. A fixed observer is inside the <wave packet> for $\Delta t=D/v_g$, and crests arrive with <wave period> $\lambda/v_p$. Thus
$$
\boxed{N_{\rm passing}=\frac{D/v_g}{\lambda/v_p}=5\frac{v_p}{v_g}=10\text{ crests}.}
$$
Crests enter the rear of the moving <wave packet> and disappear at its front, so the five visible at one instant are not five permanently identifiable crests. The count uses the intended narrow-band, approximately unchanged <wave packet> and neglects the precise visibility of its end crests.
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