A students has to put for 2 subjects out of 5 subjects for a course, namely, Commerce, Economics, Statistics, Mathematics I and Mathematics II. Mathematics II can be offered only if Mathematics I is also opted. The number of different combination of two subjects which can be opted is

Correct Option (c)

The available subjects are Commerce (C), Economics (E), Statistics (S), Mathematics I (MI), and Mathematics II (MII).

The condition states that Mathematics II (MII) can be opted only if Mathematics I (MI) is also opted.

We need to select 2 subjects. This can be analyzed in two mutually exclusive cases:

• Case 1: Mathematics II (MII) is NOT chosen.

  If MII is not chosen, the selection must be made from the remaining 4 subjects: C, E, S, and MI.

  The number of ways to choose 2 subjects from these 4 is given by the combination formula C(n, k) = n! / (k! * (n-k)!):

  C(4, 2) = 4! / (2! * (4-2)!) = 4! / (2! * 2!) = (4 × 3 × 2 × 1) / ((2 × 1) × (2 × 1)) = 12 / 2 = 6.

  The combinations are: (C, E), (C, S), (C, MI), (E, S), (E, MI), (S, MI).

• Case 2: Mathematics II (MII) IS chosen.

  According to the given condition, if MII is chosen, then MI must also be chosen.

  Since we are selecting exactly 2 subjects, this implies that the two subjects chosen must be (MI, MII).

  There is only 1 such combination: (MI, MII).

The total number of different combinations of two subjects is the sum of combinations from Case 1 and Case 2:

Total combinations = 6 + 1 = 7.

Incorrect Options:

• Option (a) 5: This count would be obtained if, for instance, Mathematics I and Mathematics II were treated as a single combined subject, or if additional restrictions were applied, which is not the case here.
• Option (b) 6: This represents only the combinations where Mathematics II is not chosen. It omits the valid combination where both Mathematics I and Mathematics II are selected.
• Option (d) 8: This value does not correspond to a logical calculation based on the given conditions. The total number of unrestricted combinations (without the MII condition) would be C(5, 2) = 10, and removing the three invalid combinations (C, MII), (E, MII), (S, MII) results in 7, not 8.