The number of strictly increasing functions \(f\) from the set \(\{1, 2, 3, 4, 5, 6\}\) to the set \(\{1, 2, 3, ...., 9\}\) such that \(f(i)>i\) for \(1 \le i \le 6\), is equal to:
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Problems involving counting ordered sequences (like strictly increasing functions) can often be transformed into simpler unordered selection problems (combinations with or without repetition). The transformation \(c_i = f(i) - i\) is very effective for converting a strict inequality \(f(i)<f(i+1)\) into a non-strict one \(c_i \le c_{i+1}\), which is the standard setup for "stars and bars".
To solve for the number of strictly increasing functions \( f \) from the set \(\{1, 2, 3, 4, 5, 6\}\) to the set \(\{1, 2, 3, \ldots, 9\}\) such that \( f(i) > i \) for \( 1 \leq i \leq 6 \), we need to follow a systematic approach:
Identify the requirement for \( f \) to be strictly increasing, and then meet the condition \( f(i) > i \).
Consider a strictly increasing function \( f : \{1, 2, 3, 4, 5, 6\} \to \{2, 3, 4, \ldots, 9\} \) since \( f(i) \) must be greater than \( i \) for each \( i \).
For \( f \) to be strictly increasing, each \( f(i) \) must be selected from the set \(\{2, 3, 4, \ldots, 9\}\) in increasing order.
Reframe the problem: We must choose 6 numbers from the set \(\{2, 3, 4, \ldots, 9\}\) (a total of 8 numbers) such that each chosen number is greater than its corresponding argument in the function.
The problem of picking 6 numbers from 8 can be solved using combinations, specifically, \(\binom{8}{6}\).
Calculate the combination:
\(\binom{8}{6} = \binom{8}{2}\) because \(\binom{n}{k} = \binom{n}{n-k}\).
