Question

_bc_ab_caabc

(a) acb (b) bab (c)aba (d) bcbb​

Answer 1

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Option- (c) aba

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Answer 2

Answer:

The series is of 12 letters.

Divide it into four groups of three letters (12/4=3).

_bc/__b/b_a/abc

Observe the logic that the series goes as 3–1–2.

abc -> cab -> bca -> abc

We get, abccabbcaabc as the final series.