If A = {1, 2, 3, 4, 5, 6}, B = {1, 2, 3, … 8, 9}. Then the number of strictly increasing functions from A → B such that f(i) ≠ i ∀ i = 1, 2, 3, 4, 5, 6 are

The correct answer is 28.

Step 1: Understand the nature of strictly increasing functions
We are given two sets:
$A=\{1,2,3,4,5,6\}$
$B=\{1,2,3,4,5,6,7,8,9\}$
Here, set A has 6 elements (n = 6) and set B has 9 elements.
A function $f:A\to B$ is strictly increasing if .
For any subset of 6 distinct elements chosen from set B, there is exactly 1 way to arrange them in strictly increasing order to form a strictly increasing function.

Step 2: Analyze the condition $f\left(i\right)\ne i$ for all $i\in \{1,2,3,4,5,6\}$
Since f is strictly increasing:
- $f\left(1\right)\ge <!--\; \ge \; -->1$
- $f\left(2\right)\ge <!--\; \ge \; -->f\left(1\right)+1\ge <!--\; \ge \; -->2$
- In general, $f\left(i\right)\ge <!--\; \ge \; -->i$ for all $i\in \{1,2,3,4,5,6\}$.
Therefore, the condition $f\left(i\right)\ne i$ is equivalent to requiring $f\left(i\right)>i$ for all i = 1, 2, 3, 4, 5, 6.
Specifically:
- $f\left(1\right)>1\Rightarrow f\left(1\right)\ge <!--\; \ge \; -->2$
- $f\left(2\right)>f\left(1\right)\ge <!--\; \ge \; -->3$
- $f\left(3\right)>f\left(2\right)\ge <!--\; \ge \; -->4$
- $f\left(4\right)>f\left(3\right)\ge <!--\; \ge \; -->5$
- $f\left(5\right)>f\left(4\right)\ge <!--\; \ge \; -->6$
- $f\left(6\right)>f\left(5\right)\ge <!--\; \ge \; -->7$

Step 3: Count the valid choices
Notice that if $f\left(1\right)\ge <!--\; \ge \; -->2$, then automatically all subsequent values satisfy $f\left(i\right)>i$ because $f\left(i\right)\ge <!--\; \ge \; -->f\left(1\right)+(i-1)\ge <!--\; \ge \; -->2+i-1=i+1>i$.
Thus, the condition $f\left(i\right)\ne i$ for all i simply means that element 1 cannot be chosen as a value in the range of f (since $f\left(1\right)\ge <!--\; \ge \; -->2$).
Therefore, the range of f must be a subset of 6 elements chosen from the set:
$B^{\prime }=\{2,3,4,5,6,7,8,9\}$
Set B' contains 9 - 1 = 8 elements.

Step 4: Compute the total number of functions
The number of ways to choose 6 distinct elements from 8 elements is given by the combination formula ${}_{8}C_6$:
${}_{8}C_6=\frac{8!}{6!\times <!--\; \times \; -->2!}=\frac{8\times <!--\; \times \; -->7}{2\times <!--\; \times \; -->1}=28$

Thus, the total number of strictly increasing functions satisfying the given condition is 28.