There are 8 students appearing in an examination of which 3 have to appear in Mathematics paper and the remaining 5 in different subjects. Then, the number of ways they can be made to sit in a row, if the candidates in Mathematics cannot sit next to each other is
1. Arrange the 5 Non-Mathematics Students: The 5 non-Mathematics students can be arranged in: \[ 5! = 120 \text{ ways} \] This creates 6 possible slots (before, between, and after the non-Mathematics students) where the 3 Mathematics students can be placed. 2. Arrange the 3 Mathematics Students: We need to choose 3 out of these 6 slots for the Mathematics students, which can be done in: \[ \binom{6}{3} = 20 \text{ ways} \] The 3 Mathematics students can be arranged in these slots in: \[ 3! = 6 \text{ ways} \] Therefore, the total number of ways to arrange the Mathematics students is: \[ 20 \times 6 = 120 \text{ ways} \] 3. Calculate the Total Number of Arrangements: Multiply the number of ways to arrange the non-Mathematics students by the number of ways to arrange the Mathematics students: \[ 120 \times 120 = 14400 \text{ ways} \] Therefore, the number of ways the students can be seated such that the Mathematics students do not sit next to each other is: \[ \boxed{14400} \]
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