Solution

Total length of bit string is $10$.

Let $A$ be the set of bit strings starting with $1$.

So, first bit is fixed and remaining $9$ bits are free.

$n(A)=2^9=512$

Let $B$ be the set of bit strings ending with $00$.

So, last two bits are fixed and remaining $8$ bits are free.

$n(B)=2^8=256$

Now, strings which start with $1$ and end with $00$:

First bit is fixed as $1$ and last two bits are fixed as $00$.

Remaining $7$ bits are free.

$n(A \cap B)=2^7=128$

Therefore,

$n(A \cup B)=n(A)+n(B)-n(A \cap B)$

$=512+256-128$

$=640$