Concept:
To arrange these Carnatic rhythmic cycles in ascending order according to their total beat counts, we first calculate the number of matras for each taal based on the Tisra Jati modification (where the variable *Laghu* block is worth exactly 3 beats).
Step 1: Calculate total beats for each option using Tisra Jati component values.
Recall that a *Dhrutam* ($O$) has a fixed length of 2 beats, an *Anudhrutam* ($U$) has a fixed length of 1 beat, and a *Tisra Laghu* ($|$) has a length of 3 beats. Let us calculate the total beat counts:
• C. Ektaal: Contains a single Laghu ($|$).
$$\text{Total beats} = 3 \text{ matras}$$
• D. Jhamptaal: Formed by one Laghu, one Anudhrutam, and one Dhrutam ($| \, U \, O$).
$$\text{Total beats} = 3 \, (\text{Laghu}) + 1 \, (\text{Anudhrutam}) + 2 \, (\text{Dhrutam}) = 6 \text{ matras}$$
• E. Triputa Taal: Formed by one Laghu followed by two Dhrutams ($| \, O \, O$).
$$\text{Total beats} = 3 \, (\text{Laghu}) + 2 \, (\text{Dhrutam}) + 2 \, (\text{Dhrutam}) = 7 \text{ matras}$$
• B. Mathtaal: Formed by one Laghu, one Dhrutam, and another Laghu ($| \, O \, |$).
$$\text{Total beats} = 3 \, (\text{Laghu}) + 2 \, (\text{Dhrutam}) + 3 \, (\text{Laghu}) = 8 \text{ matras}$$
• A. Dhruvtaal: Formed by one Laghu, one Dhrutam, and two consecutive Laghus ($| \, O \, | \, |$).
$$\text{Total beats} = 3 + 2 + 3 + 3 = 11 \text{ matras}$$
Step 2: Arrange the totals in ascending order.
Ordering these calculated values from lowest to highest ($3 < 6 < 7 < 8 < 11$) produces the following sequence: Ektaal (C) \(\rightarrow\) Jhamptaal (D) \(\rightarrow\) Triputa (E) \(\rightarrow\) Mathtaal (B) \(\rightarrow\) Dhruvtaal (A). This sequence corresponds to option (1).