Study the following information carefully and answer the question given below

Eight persons from A to H has different degree i.e., BBA, MBA and MCA but not necessarily in the same order. At least two but not more than three persons have the same degree. Consecutive alphabetically named persons do not have the same degree.

D does not have an MBA degree. Only C and F have the same degree but not a BBA degree. Both A and H have the same degree but not as D have. E does not have an MBA degree. B does not have a BBA degree.

Which of the following pair/group of combination(s) is/are correct?

Correct Answer: H- BBA

Step-by-step Explanation:

Let's analyze the given clues step-by-step to deduce the degree for each of the eight persons (A, B, C, D, E, F, G, H):

1. Basic Rules & Constraints:

• Degrees available: BBA, MBA, MCA.

• At least 2 and at most 3 persons have the same degree.

• Consecutive alphabetically named persons (e.g., A and B, B and C, C and D, etc.) cannot have the same degree.

2. Analyzing Clues:

• "Only C and F have the same degree but not a BBA degree."
Since "only" C and F share their degree, this specific degree group consists of exactly 2 persons. The available degrees are BBA, MBA, and MCA. Since C and F do not have BBA, they must have either MBA or MCA. Let's determine which one: if C and F had MBA, no one else could have MBA. But if 2 persons have MBA, then the remaining 6 persons must be divided between BBA and MCA, which would mean 3 in BBA and 3 in MCA (total 3+3+2 = 8).
Wait, let's look at C and F: C and F are not alphabetically consecutive (C-D-E-F), so they can share a degree.

• "D does not have an MBA degree." and "E does not have an MBA degree."
Since D and E are consecutive alphabetically, D and E cannot have the same degree. Since neither has MBA, one of them must have BBA and the other must have MCA.

• "Both A and H have the same degree but not as D have."
So A and H have the same degree.

• "B does not have a BBA degree."

3. Deduction of Degree Groups:

Since C and F are the ONLY two persons having their degree, that degree has exactly 2 members.
The other two degrees must each be shared by 3 persons (since 2 + 3 + 3 = 8 total persons, respecting the max 3 per degree rule).

Let's test the degree for C and F:
If C and F have MCA (2 persons):
Then BBA must have 3 persons, and MBA must have 3 persons.

Let's check the constraints:
• A and H have the same degree.
• Since C and F are the ONLY ones with MCA, A and H cannot have MCA.
• Therefore, A and H must have either BBA or MBA.
• Given the correct option is H- BBA, A and H both have BBA.
• Since D has a different degree from A and H, D cannot have BBA. Since D also does not have MBA, D must have MCA... wait, but C and F are the ONLY ones with MCA. Hence, MCA cannot be the degree with only 2 members if D has MCA.
Instead, let's re-evaluate:

If C and F have MCA (2 persons): C- MCA, F- MCA.
Then BBA (3 persons) and MBA (3 persons).
• A and H share a degree, so A and H are in BBA (since they can't be in MCA, and let's check BBA vs MBA).
• If A and H have BBA, then the third person in BBA must be chosen such that no consecutive letters match.
• Since D does not have BBA (as D's degree is different from A and H), D has MCA? No, C and F are the ONLY ones in MCA. So D must have MBA! But the clue says "D does not have an MBA degree."
Therefore, C and F must have MBA (2 persons): C- MBA, F- MBA!

Let's verify C and F having MBA (2 persons):
• Degrees distribution: MBA (2 persons: C, F), BBA (3 persons), MCA (3 persons).
• D does not have MBA, E does not have MBA.
• A and H have the same degree, which is NOT D's degree.
• If A and H have BBA, then D (which has a different degree from A and H) must have MCA!
- Is D having MCA valid? Yes, because MCA has 3 persons (not limited to only 2).
- Does D have MBA? No (matches clue).
• E does not have MBA, and E is adjacent to D, so E cannot have MCA (since D has MCA). Thus, E must have BBA!
• Now BBA has: A, H, E (3 persons).
- Let's check adjacent letters in BBA: A, E, H. None are consecutive! (A-B, D-E-F, G-H). Perfect!
• MCA has 3 persons: D is one of them.
• B does not have BBA. B cannot have MBA (since only C and F have MBA). Thus, B must have MCA!
• G: BBA is full (A, E, H). MBA is full (C, F). So G must have MCA!
• Now MCA has: B, D, G (3 persons).
- Let's check adjacent letters in MCA: B, D, G. None are consecutive! Perfect!

4. Final Combination Summary:

• BBA (3 persons): A, E, H
• MBA (2 persons): C, F
• MCA (3 persons): B, D, G

Checking the given options:
• B- MBA (Incorrect, B has MCA)
• C- MCA (Incorrect, C has MBA)
• H- BBA (Correct!)
• B, E - MCA (Incorrect, E has BBA)
• None is correct (Incorrect)

Thus, the correct combination is H- BBA.