How to solve this can anybody explain

https://youtu.be/9kFN4gw9cQI

you gotta study the clock darling

Setup — measure each hand as a fraction of the full circle from 12.

	position
hour hand	$x$
minute hand	$12x$

Swap them:

	position after swap	what it must be
hour hand	$12x$	$12x$
minute hand	$x$	$12(12x)=144x$

So we need $144x$ and $x$ to be the same point on the dial:

$144x - x = 143x = \text{whole number}$

Count:

set	condition	count
valid swaps	$143x \in \mathbb{Z}$	$x=\tfrac{0}{143},\dots,\tfrac{142}{143}$ → 143
hands coincide	$12x-x=11x \in \mathbb{Z}$	$x=\tfrac{0}{11},\dots,\tfrac{10}{11}$ → 11

Coincidences sit inside the 143 (since $\frac{j}{11}=\frac{13j}{143}$), so subtract:

$143-11=\boxed{132}$