The number of strictly increasing functions f$f$ from the set \1, 2, 3, 4, 5, 6\$\{1, 2, 3, 4, 5, 6\}$ to the set \1, 2, 3, ldots, 9\$\{1, 2, 3, \ldots, 9\}$ such that f(i) neq i$f(i) \neq i$ for 1 leq i leq 6$1 \leq i \leq 6$ , is equal to:

### Related Formula For a strictly increasing function f: A to B$f: A \to B$ where |A| = m$|A| = m$ and |B| = n$|B| = n$, the number of functions without restrictions is binomnm$\binom{n}{m}$. ### Core Logic We need strictly increasing functions f: \1,2,3,4,5,6\ to \1,2,dots,9\$f: \{1,2,3,4,5,6\} \to \{1,2,\dots,9\}$ subject to f(i) neq i$f(i) \neq i$. Since f$f$ is strictly increasing, f(i) geq i$f(i) \geq i$ must always hold because the target values are drawn from an equally spaced domain. If f(i) = i$f(i) = i$ for any i$i$, it forces a strict ladder down to 1. But we are given f(i) neq i$f(i) \neq i$. Thus, f(i) > i$f(i) > i$ for all 1 leq i leq 6$1 \leq i \leq 6$. This implies f(1) geq 2$f(1) \geq 2$.

### Step 1: Case Analysis on f(1) Since f(1) > 1$f(1) > 1$, we evaluate possible starting points: Case 1: f(1) = 2$f(1) = 2$ Remaining 5 values f(2)$f(2)$ to f(6)$f(6)$ must be strictly increasing and chosen from \3, 4, 5, 6, 7, 8, 9\$\{3, 4, 5, 6, 7, 8, 9\}$ (7 available numbers). Since f(1)=2$f(1)=2$, f(i)>i$f(i)>i$ is naturally preserved for subsequent elements (e.g., f(2) ge 3 > 2$f(2) \ge 3 > 2$). Number of ways = binom75 = 21$= \binom{7}{5} = 21$.

### Step 2: Subsequent Cases Case 2: f(1) = 3$f(1) = 3$ Remaining 5 values chosen from \4, 5, 6, 7, 8, 9\$\{4, 5, 6, 7, 8, 9\}$ (6 available numbers). Number of ways = binom65 = 6$= \binom{6}{5} = 6$. Case 3: f(1) = 4$f(1) = 4$ Remaining 5 values chosen from \5, 6, 7, 8, 9\$\{5, 6, 7, 8, 9\}$ (5 available numbers). Number of ways = binom55 = 1$= \binom{5}{5} = 1$. Case 4: f(1) = 5$f(1) = 5$ Requires choosing 5 values from \6,7,8,9\$\{6,7,8,9\}$, which is impossible. ### Step 3: Total Sum Total number of valid functions = 21 + 6 + 1 = 28$21 + 6 + 1 = 28$. ### Pattern Recognition For f(i) neq i$f(i) \neq i$ on strictly increasing integer arrays, f(x) - x > 0$f(x) - x > 0$. Using the substitution g(x) = f(x) - x$g(x) = f(x) - x$, you convert a constrained increasing function into a standard non-decreasing one, or simply pivot on f(1)$f(1)$ and sum the cascading binomials. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Permutations and Combinations Class 12 Maths: Functions