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(B) To solve this,we look for a repeating pattern in the series: $ab...bbc...c...ab...ab...b$.
Let's count the total number of characters includi

Vedclass · Accepted Answer

$cbabc$

Vedclass · Answer

(B) To solve this,we look for a repeating pattern in the series: $ab...bbc...c...ab...ab...b$.
Let's count the total number of characters including blanks: $18$.
We can divide the series into groups of $3$ or $6$.
If we divide it into groups of $3$: $(ab.), (bbc), (.c.), (ab.), (ab.), (.b)$. This does not seem consistent.
Let's try groups of $6$: $(ab.bbc), (c...ab...), (ab...b)$.
Looking at the pattern $abcbbc$,let's test the sequence $abcbbc$ repeating:
$ab(c)bbc$ | $c(a)bc(a)b$ | $c(a)b(c)ab(c)b$.
Actually,let's try filling the blanks with $c, c, a, a, c$:
$ab(c)bbc | c(c)c(a)ab | (a)b(c)ab(c)b$.
Wait,let's re-examine the pattern: $abc, bbc, cab, cab, abc, b$. This is not it.
Let's try the pattern $abc, bbc$ repeating:
$ab(c)bbc | c(a)bc(a)c | ab(c)ab(c)b$.
Correct pattern is $abc, bbc$ repeating: $ab(c)bbc | c(a)bc(a)c | ab(c)ab(c)b$ is not correct.
Let's try option $A$ $(ccaac)$: $ab(c)bbc | c(c)c(a)ab | (a)b(c)ab(c)b$ - No.
Let's try option $B$ $(cbabc)$: $ab(c)bbc | c(b)c(a)bc | (b)ab(c)ab(c)b$ - No.
Let's try option $C$ $(cacac)$: $ab(c)bbc | c(a)c(c)ab | (a)b(c)ab(c)b$ - No.
Let's try the pattern $abc, bbc$ again. The sequence is $ab(c)bbc | c(a)bc(a)c | ab(c)ab(c)b$. Actually,the pattern is $abc, bbc$ repeated: $ab(c)bbc, c(a)bc(a)c, ab(c)ab(c)b$. The correct sequence is $abc, bbc, cab, cab, abc, bbc$. Filling $c, b, a, b, c$ gives $ab(c)bbc | c(b)c(a)bc | (b)ab(c)ab(c)b$.

Choose the correct alternative to fill in the blanks in the series: $ab...bbc...c...ab...ab...b$

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