We are given two sets $A = \{1, 2, 3\}$ and $B = \{a, b, c, d, e\}$. We need to solve the following problems: i) Find the total number of functions from $A$ to $B$. ii) Find the number of injective functions from $A$ to $B$. iii) Find the number of injective functions from $A$ to $B$ such that $f(1) = a$. iv) Find the number of injective functions from $A$ to $B$ such that $f(1) \ne a$ and $f(2) \ne b$.

1. Problem Description

We are given two sets $A = \{1, 2, 3\}$ and $B = \{a, b, c, d, e\}$. We need to solve the following problems:

i) Find the total number of functions from $A$ to $B$.

ii) Find the number of injective functions from $A$ to $B$.

iii) Find the number of injective functions from $A$ to $B$ such that $f(1) = a$.

iv) Find the number of injective functions from $A$ to $B$ such that $f(1) \ne a$ and $f(2) \ne b$.

2. Solution Steps

i) The total number of functions from a set $A$ with $|A| = m$ to a set $B$ with $|B| = n$ is $n^m$. Here, $|A| = 3$ and $|B| = 5$, so the number of functions from $A$ to $B$ is $5^3 = 125$.

ii) The number of injective functions from a set $A$ with $|A| = m$ to a set $B$ with $|B| = n$ where $m \le n$ is given by the formula:

$P(n, m) = \frac{n!}{(n-m)!}$.

In our case, $n = 5$ and $m = 3$. So, the number of injective functions is:

$P(5, 3) = \frac{5!}{(5-3)!} = \frac{5!}{2!} = \frac{120}{2} = 60$.

iii) If $f(1) = a$, then we need to find the number of injective functions from the remaining two elements of $A$, $\{2, 3\}$, to the remaining four elements of $B$, $\{b, c, d, e\}$. The number of such injective functions is $P(4, 2) = \frac{4!}{(4-2)!} = \frac{4!}{2!} = \frac{24}{2} = 12$.

iv) We need to find the number of injective functions from $A$ to $B$ such that $f(1) \ne a$ and $f(2) \ne b$.

First, we compute the total number of injective functions from $A$ to $B$, which is $60$.

We subtract the number of injective functions such that $f(1) = a$, which is $12$.

We subtract the number of injective functions such that $f(2) = b$. If $f(2) = b$, then we map the remaining two elements of $A$ to the remaining four elements of $B$. The number of such injective functions is $P(4, 2) = \frac{4!}{2!} = 12$.

However, we subtracted twice the number of injective functions such that $f(1) = a$ and $f(2) = b$. If $f(1) = a$ and $f(2) = b$, then we need to find an injective function from the remaining one element of $A$, $\{3\}$, to the remaining three elements of $B$, $\{c, d, e\}$. The number of such injective functions is $P(3, 1) = \frac{3!}{(3-1)!} = \frac{3!}{2!} = 3$.

Using inclusion-exclusion, we have:

Total injective functions = 60

$f(1) = a$: 12

$f(2) = b$: 12

$f(1) = a$ and $f(2) = b$: 3

The number of injective functions where $f(1) \ne a$ and $f(2) \ne b$ is:

$60 - 12 - 12 + 3 = 39$

Another way to solve this is:

There are 5 choices for $f(1)$. Since $f(1) \ne a$, there are 4 choices for $f(1)$.

Now consider $f(2)$. There are 5 choices for where $f(2)$ can be mapped. But $f(2) \ne f(1)$ and $f(2) \ne b$.

If $f(1) = b$, then there are 4 choices for $f(2)$ since $f(2) \ne b$ and $f(2) \ne f(1)$.

If $f(1) \ne b$, then there are 3 choices for $f(2)$ since $f(2) \ne b$ and $f(2) \ne f(1)$.

Then we have two cases.

Case 1: $f(1) = b$. In this case, we have 1 choice for $f(1)$ and 4 choices for $f(2)$. Then there are 3 choices for $f(3)$, for a total of $1 \times 4 \times 3 = 12$ functions.

Case 2: $f(1) \ne b$ and $f(1) \ne a$. In this case, we have 3 choices for $f(1)$, 3 choices for $f(2)$ since $f(2) \ne b$ and $f(2) \ne f(1)$. Then there are 3 choices for $f(3)$, for a total of $3 \times 3 \times 3 = 27$ functions.

So, the total number of such injective functions is $12 + 27 = 39$.

3. Final Answer

i) 125

ii) 60

iii) 12

iv) 39