Distribute $10$ distinct prizes to $4$ students

Question

Inclusion Exclusion Used

Theorem 8.1

The Question:

In how many ways can one distribute $10$ distinct prizes to $4$ students with exactly $2$ students getting nothing. b) How many ways have at least 2 students receiving prizes?

My Work:

Let $S =$ {All ways to distribute the prizes}

$|S|= 10^4$

$A =$ student 1 gets a prize

$B =$ student 2 gets a prize

$C =$ student 3 gets a prize

$D =$ student 4 gets a prize

By the Inclusion-Exclusion** formula given, our answer is:

$S_2 - \binom{3}{1}S_3 + \binom{4}{2}S_4$

$S_2 = 100*6 =$ all distinct unions between 2 sets summed together

$S_3 = 1000*4$ all distinct unions between 3 of our sets summed together

$600-4*1000*4+C(4,2)*10000 = 44'600$

My Question

I don't think how I'm doing this question is right because I got the wrong answer from the back of my book. I'm wondering where I went wrong.

Answer 1

Is the answer 6132?

${4\choose 2}=6$ ways to pick the two students who get a prize. Then the 10 prizes can each go to one of those two students, with the exceptions of the two cases where one student gets all of the prizes.

So $6\times1022=6132$?