A class consists of 15 boys of whom 5 are prefects. How many committees of 8 can be formed if each consists of at least 2 prefects? A class consists of $15$ boys of whom $5$ are prefects. How many committees of 8 can be formed if each consists of at least $2$ prefects?

I had the solution following solution in mind but that seems to be incorrect. can anyone please explain to me why?

First, I select $2$ students from the perfect category using $\binom{5}{2}$ and then I select remaining $6$ students combined from both the perfect and unperfect category i.e. $\binom{13}{6}$. So the answer becomes $\binom{5}{2}\times \binom{13}{6} = 17160$.
But, the correct answer is $5790$. Can you please help me?
There should be minimum $2$ perfects in the committee, there is no limit for maximum of perfects. So, The number of ways can be counted as:

$1$. $2$ perfect and $6$ non perfect..... Number of ways become ${{5}\choose{2}}\times{{10}\choose{6}}$.

$2$. $3$ perfect and $5$ non perfect....... Number of ways become ${{5}\choose{3}}\times{{10}\choose{5}}$.

$3$. $4$ perfect and $4$ non perfect.....Number of ways become ${{5}\choose{4}}\times{{10}\choose{4}}$.

$4$.  $5$ perfect and $3$ non perfect.....Number of ways become ${{5}\choose{5}}\times{{10}\choose{3}}$.

Add all the ways and get the answer.
15C8-(10C8 + 10C7*5C1) = 5790

At least two means, total number of ways - number of ways to make committee without any prefect + number of ways to make committee with one prefect and 7 non prefects.

15C8 - total number of ways to choose 8 people committee.
10C8 - number of ways not to choose any prefects.
10C7*5C1 - number of ways to choose one prefect and 7 non prefects