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 $f(i)
eq i$ for $1 \leq i \leq 6$, is equal to :

Question

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 $f(i) 
eq i$ for $1 \leq i \leq 6$, is equal to :

Accepted Answer

B. 1. **Understand the setup.** A strictly increasing function $f:\{1,\dots,6\}	o\{1,\dots,9\}$ is completely determined by choosing $6$ values (the images) from $\{1,\dots,9\}$ and placing them in increasing order. So counting valid functions reduces to counting valid $6$-element subsets satisfying $f(i)
eq i$.

2. **Split into cases based on $f(1)$.** Since $f$ is strictly increasing, once $f(1)$ is fixed large enough, all later values automatically avoid the clash $f(i)=i$.

3. **Case $f(1)=2$:** The remaining $5$ images are chosen from $\{3,4,\dots,9\}$ (7 values), and being placed in increasing order automatically keeps $f(i)
eq i$ for all $i$.
$$\binom{7}{5} = 21$$

4. **Case $f(1)=3$:** The remaining $5$ images are chosen from $\{4,5,\dots,9\}$ (6 values).
$$\binom{6}{5} = 6$$

5. **Case $f(1)=4$:** The remaining $5$ images are chosen from $\{5,6,\dots,9\}$ (5 values).
$$\binom{5}{5} = 1$$

6. **Add all the cases together.**
$$21+6+1 = 28$$

Hence, the answer is **Option B (28)**.

The number of strictly increasing functions fff from the set {1,2,3,4,5,6}\{1,2,3,4,5,6\}{1,2,3,4,5,6} to the set {1,2,3,…,9}\{1,2,3, \ldots, 9\}{1,2,3,…,9} such that f(i)≠if(i) \neq if(i)=i for 1≤i≤61 \leq i \leq 61≤i≤6, is equal to :

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 $f(i) \neq i$ for $1 \leq i \leq 6$ , is equal to :