In a class of 200 students,70 offered physics,90 chemistry,100 mathematics while 24 did not offer any of the three subject,23 student offered physics and chemistry,41 chemistry and mathematics while 8 offered all the three subjects.How many students offered exactly two of the subjects?

Let P be the number of students offered Physics (P=70),

C be the number of students offered Chemistry (C=90),

M be the number of students offered Mathematics (M=100),

PC be the number of students offered Physics and Chemistry (PC=23),

CM be the number of students offered Chemistry and Mathematics (CM = 41),

PM be the number of students offered Physics and Mathematics (PM = ?),

PCM be the number of students offered Physics, Chemistry and Mathematics (PCM = 8),

P|C|M be the number of students offered at least one of three subjects.

P|C|M = 200 - 24 = 176.

Using the inclusion-exclusion principle, we have:

P|C|M = P + C + M - PC - PM - CM +PCM,

from where

PM = P + C + M - PC - CM +PCM - P|C|M = 70 + 90 + 100 - 23 - 41 + 8 - 176 = 28.

Let now $PC\bar M$ be the number of students offered Physics and Chemistry but not Mathematics, $P\bar CM$ be the number of students offered Physics and Mathematics but not Chemistry, and $\bar PCM$ be the number of students offered Mathematics and Chemistry but not Physics. Then

$PC\bar M = PC - PCM = 23 - 8 = 15$

$P\bar CM = PM - PCM = 28 - 8 = 20$$\bar PCM = CM - PCM = 41 - 8 = 33$

The total number of students offered exactly 2 subjects is equal to

$PC\bar M + P\bar CM + \bar PCM = 15 + 20 + 33 = 68$

Answer. 68 students offered exactly 2 subjects.