$(7)(2$ pts$)$ In the routes$-$aggregation example below, it is said that using the maximum sequencing matching rule,

router A can still perform routes aggregation by combining the three networks into one and report

$200\mathrm{.}23\mathrm{.}00010000\mathrm{.}0/22$ to router B$.$ D will report $200\mathrm{.}23\mathrm{.}00010011\mathrm{.}0/24$ to B$.$ Thus there should be two

route entries in B$\text{'}$s routing table.

Someone has raised the following question. If a packet arrives at B and belongs to the top group.

Assume it has the following destination address: $200\mathrm{.}23\mathrm{.}00010011\mathrm{.}234.$ But according to the maximum

sequencing rule, we will send it to port $2.$ This leads to a routing mistake. Does this question make sense?

If so$,$ how to fix the problem. If not, why not.

$200\mathrm{.}23\mathrm{.}00010000\mathrm{.}0$

$200\mathrm{.}23\mathrm{.}00010001\mathrm{.}0$

$200\mathrm{.}23\mathrm{.}00010010\mathrm{.}0$

$200\mathrm{.}23\mathrm{.}00010011\mathrm{.}0$