$5$ chem students, $6$ maths students and $7$ physics students permutation $5$ chem students, $6$ maths students and $7$ physics students.
Find the number of arrangements if
a)Chem majors are to occupy the first 5 positions
b)Chem majors cannot occupy the first 5 positions
c)Students with the same major must be together in a block
a)$5!13!$
b)$18! - 5!13!$
c)$5!6!7! \times 3!$
Is my part B wrong??
Answer sheet for part B says:
First the chemistry majors must be place among the last $13$ positions
($P(13, 5)$ ways). Then place the other majors ($13!$ ways). Thus ans: $P(13, 5)13! = \frac{(13!)^2}{8!}$.
I don't get it though.
 A: It is a matter of interpretation. The author interpreted the question as meaning we cannot have any Chem major among the first five. 
You counted the number of ways in which not all of the first five are Chem majors, though some might be.  For you subtracted the number of ways all of the first five positions are occupied by Chem majors from the total number of permutations.
Remark: I would actually lean towards your interpretation. If the author wished to exclude Chem students from the first five positions, b) should have said Chem majors cannot occupy any of the first five positions.
A: For part a fix $5$ chem major in the first five positions and trat them as a block. Then the total number number of ways rest majors are permutated is $(6+7)!=13!$ and the number of ways in which the five chem majors permutate in their  positions is $5!$. Then apply multiplication rule of permutation.
For part b first calculate total number of ways 18 different majors are arranged  in $18$ different places is $(5+6+7)!=18!$ and subtract the number of ways in which $5$ five chem major sat in in the first five positions.
For part c treat each major a block. 
