Counting strictly increasing triplets from 10 ordered positives: Given 10 positive real numbers n1 < n2 < n3 < … < n10, how many distinct triplets (ni, nj, nk) with i < j < k can be formed (i.e., strictly increasing triples by index)?

Introduction / Context:
This is a combinations question: each strictly increasing triple corresponds to a set of three distinct indices i < j < k chosen from 10 positions. The actual values do not affect the count because the ordering is already strict by index.

Given Data / Assumptions:

• Sequence of 10 numbers with n1 < n2 < … < n10.
• Triplet must respect i < j < k (strictly increasing by index).

Concept / Approach:

• Choose 3 indices out of 10: combinations without order.
• Each choice yields exactly one increasing triplet.

Step-by-Step Solution:

Verification / Alternative check:
Any attempt to count by “sliding windows” would undercount (e.g., 8). The correct method uses combinations of indices, not adjacency of terms.

Why Other Options Are Wrong:

• 45 and 90 are incorrect binomial coefficients for other parameters.
• 180 counts some permutations or misapplies arrangements.

Common Pitfalls:

• Confusing “consecutive triplets” with “any strictly increasing triplets.”

Final Answer:
120