Competency-Based
Assessment Made Simple.

Step-by-Step Solution

1. First, we find the total number of strictly increasing functions from $A$ to $B$ without any restrictions. Since $A$ has 6 elements and $B$ has 9 elements, we need to choose 6 distinct elements from $B$ and arrange them in increasing order. The number of ways to do this is $\binom{9}{6} = \frac{9!}{6!3!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.

2. Now, we use the Principle of Inclusion-Exclusion to account for the condition $f(i) \ne i$ for all $i$. Let $S_i$ be the set of functions such that $f(i) = i$. We want to find the number of functions such that none of the $f(i) = i$ holds.

3. • $|S_1|$: If $f(1) = 1$, we need to choose 5 elements from the remaining 8 elements in $B$ such that the function is strictly increasing. This is $\binom{8}{5} = 56$.
   • $|S_1 \cap S_2|$: If $f(1) = 1$ and $f(2) = 2$, we need to choose 4 elements from the remaining 7 elements in $B$. This is $\binom{7}{4} = 35$.
   • $|S_1 \cap S_2 \cap S_3|$: If $f(1) = 1$, $f(2) = 2$, and $f(3) = 3$, we need to choose 3 elements from the remaining 6 elements in $B$. This is $\binom{6}{3} = 20$.
   • $|S_1 \cap S_2 \cap S_3 \cap S_4|$: If $f(1) = 1$, $f(2) = 2$, $f(3) = 3$, and $f(4) = 4$, we need to choose 2 elements from the remaining 5 elements in $B$. This is $\binom{5}{2} = 10$.
   • $|S_1 \cap S_2 \cap S_3 \cap S_4 \cap S_5|$: If $f(1) = 1$, $f(2) = 2$, $f(3) = 3$, $f(4) = 4$, and $f(5) = 5$, we need to choose 1 element from the remaining 4 elements in $B$. This is $\binom{4}{1} = 4$.
   • $|S_1 \cap S_2 \cap S_3 \cap S_4 \cap S_5 \cap S_6|$: If $f(1) = 1$, $f(2) = 2$, $f(3) = 3$, $f(4) = 4$, $f(5) = 5$, and $f(6) = 6$, we need to choose 0 elements from the remaining 3 elements in $B$. This is $\binom{3}{0} = 1$.

4. Using the Principle of Inclusion-Exclusion, the number of functions such that $f(i) \ne i$ for all $i$ is:

   $\binom{9}{6} - \binom{6}{1}\binom{8}{5} + \binom{6}{2}\binom{7}{4} - \binom{6}{3}\binom{6}{3} + \binom{6}{4}\binom{5}{2} - \binom{6}{5}\binom{4}{1} + \binom{6}{6}\binom{3}{0}$

   $= 84 - 6(56) + 15(35) - 20(20) + 15(10) - 6(4) + 1(1)$

   $= 84 - 336 + 525 - 400 + 150 - 24 + 1 = 84 + 525 + 150 + 1 - 336 - 400 - 24 = 760 - 760 = 0$

5. The calculation above is incorrect. Let's recalculate using the Principle of Inclusion-Exclusion.

   $N = \binom{9}{6} - \sum_{i=1}^6 (-1)^{i-1} \binom{6}{i} \binom{9-i}{6-i}$

   $N = \binom{9}{6} - \binom{6}{1}\binom{8}{5} + \binom{6}{2}\binom{7}{4} - \binom{6}{3}\binom{6}{3} + \binom{6}{4}\binom{5}{2} - \binom{6}{5}\binom{4}{1} + \binom{6}{6}\binom{3}{0}$

   $N = 84 - 6(56) + 15(35) - 20(20) + 15(10) - 6(4) + 1(1)$

   $N = 84 - 336 + 525 - 400 + 150 - 24 + 1 = 0$

6. Let $S$ be the set of all strictly increasing functions from $A$ to $B$. Then $|S| = \binom{9}{6} = 84$. Let $A_i$ be the set of functions in $S$ such that $f(i) = i$. We want to find $|S \setminus (A_1 \cup A_2 \cup \dots \cup A_6)|$.

   By the Principle of Inclusion-Exclusion:

   $|A_1 \cup A_2 \cup \dots \cup A_6| = \sum |A_i| - \sum |A_i \cap A_j| + \sum |A_i \cap A_j \cap A_k| - \dots + (-1)^{6-1} |A_1 \cap A_2 \cap \dots \cap A_6|$

   $|A_i| = \binom{9-1}{6-1} = \binom{8}{5} = 56$. There are $\binom{6}{1}$ such terms.

   $|A_i \cap A_j| = \binom{9-2}{6-2} = \binom{7}{4} = 35$. There are $\binom{6}{2}$ such terms.

   $|A_i \cap A_j \cap A_k| = \binom{9-3}{6-3} = \binom{6}{3} = 20$. There are $\binom{6}{3}$ such terms.

   $|A_i \cap A_j \cap A_k \cap A_l| = \binom{9-4}{6-4} = \binom{5}{2} = 10$. There are $\binom{6}{4}$ such terms.

   $|A_i \cap A_j \cap A_k \cap A_l \cap A_m| = \binom{9-5}{6-5} = \binom{4}{1} = 4$. There are $\binom{6}{5}$ such terms.

   $|A_1 \cap A_2 \cap A_3 \cap A_4 \cap A_5 \cap A_6| = \binom{9-6}{6-6} = \binom{3}{0} = 1$. There is $\binom{6}{6}$ such term.

   $|A_1 \cup A_2 \cup \dots \cup A_6| = \binom{6}{1}\binom{8}{5} - \binom{6}{2}\binom{7}{4} + \binom{6}{3}\binom{6}{3} - \binom{6}{4}\binom{5}{2} + \binom{6}{5}\binom{4}{1} - \binom{6}{6}\binom{3}{0}$

   $= 6(56) - 15(35) + 20(20) - 15(10) + 6(4) - 1(1) = 336 - 525 + 400 - 150 + 24 - 1 = 860 - 676 = 84$

   So, $|S \setminus (A_1 \cup A_2 \cup \dots \cup A_6)| = |S| - |A_1 \cup A_2 \cup \dots \cup A_6| = 84 - 84 = 0$