Question:

Step 1: Calculate the total number of injective mappings from $A$ to $B$. The set $A$ has 4 elements and the set $B$ has 6 elements. An injective mapping selects 4 distinct elements from $B$ and assigns them to the 4 elements of $A$ in order. The total number of injective mappings is given by the permutation formula $P(n, k) = \frac{n!}{(n-k)!}$. $$P(6,4) = \frac{6!}{(6-4)!} = \frac{6!}{2!} = 6 \times 5 \times 4 \times 3 = 360$$ Step 2: Apply the Principle of Inclusion-Exclusion. We need to find the number of injective mappings $f: A \to B$ such that $f(i) \neq i$ for all $i \in A$. Let $S_i$ be the set of injective mappings where $f(i)=i$. We want to find the number of mappings that are not in $\bigcup_{i=1}^4 S_i$. This is given by the total number of injective mappings minus the number of mappings in $\bigcup_{i=1}^4 S_i$. By the Principle of Inclusion-Exclusion: $$ \left| \bigcup_{i=1}^4 S_i \right| = \sum |S_i| - \sum |S_i \cap S_j| + \sum |S_i \cap S_j \cap S_k| - |S_1 \cap S_2 \cap S_3 \cap S_4| $$ Step 3: Calculate $\sum |S_i|$. For any $i \in A$, if $f(i)=i$, then one element of $A$ is fixed. We need to map the remaining $4-1=3$ elements of $A$ to the remaining $6-1=5$ elements of $B$. The number of ways to do this is $P(5,3) = 5 \times 4 \times 3 = 60$. There are $\binom{4}{1}=4$ such sets $S_i$. $$ \sum |S_i| = \binom{4}{1} \times P(5,3) = 4 \times 60 = 240 $$ Step 4: Calculate $\sum |S_i \cap S_j|$. For any distinct $i, j \in A$, if $f(i)=i$ and $f(j)=j$, then two elements of $A$ are fixed. We need to map the remaining $4-2=2$ elements of $A$ to the remaining $6-2=4$ elements of $B$. The number of ways to do this is $P(4,2) = 4 \times 3 = 12$. There are $\binom{4}{2}=6$ such pairs of sets $S_i \cap S_j$. $$ \sum |S_i \cap S_j| = \binom{4}{2} \times P(4,2) = 6 \times 12 = 72 $$ Step 5: Calculate $\sum |S_i \cap S_j \cap S_k|$. For any distinct $i, j, k \in A$, if $f(i)=i, f(j)=j, f(k)=k$, then three elements of $A$ are fixed. We need to map the remaining $4-3=1$ element of $A$ to the remaining $6-3=3$ elements of $B$. The number of ways to do this is $P(3,1) = 3$. There are $\binom{4}{3}=4$ such triples of sets $S_i \cap S_j \cap S_k$. $$ \sum |S_i \cap S_j \cap S_k| = \binom{4}{3} \times P(3,1) = 4 \times 3 = 12 $$ Step 6: Calculate $|S_1 \cap S_2 \cap S_3 \cap S_4|$. If $f(1)=1, f(2)=2, f(3)=3, f(4)=4$, then all four elements of $A$ are fixed. We need to map the remaining $4-4=0$ elements of $A$ to the remaining $6-4=2$ elements of $B$. The number of ways to do this is $P(2,0) = 1$. There is $\binom{4}{4}=1$ such quadruple of sets $S_1 \cap S_2 \cap S_3 \cap S_4$. $$ |S_1 \cap S_2 \cap S_3 \cap S_4| = \binom{4}{4} \times P(2,0) = 1 \times 1 = 1 $$ Step 7: Calculate the number of mappings satisfying the condition. The number of mappings where at least one $f(i)=i$ is given by the Principle of Inclusion-Exclusion. $$ \left| \bigcup_{i=1}^4 S_i \right| = 240 - 72 + 12 - 1 = 179 $$ The number of injective mappings satisfying $f(i) \neq i$ for all $i \in A$ is the total number of injective mappings minus the number of mappings where at least one $f(i)=i$. $$ 360 - 179 = 181 $$