Injective functions between sets Given the statement: 
The number of injective functions $f:\{1,2,3,4\} \to \{1,2,3,4,5\}$ such that $\{1,2,3\} \subseteq f[\{1,2,3,4\}]$ equals to the number of The number of injective functions $f:\{1,2,3,4\} \to \{1,2,3,4,5\}$ such that $\{1,2\} \not\subseteq f[\{1,2,3,4\}]$
My question is:
The statement is true or false?

MY APPROACH: First, I calculated the number of injective
functions such that $\{1,2,3\} \subseteq f[\{1,2,3,4\}]$ and I got
$3! \cdot 5=30$ Second, I calculated the number of all functions
available from $A$ to $B$: $f:\{1,2,3,4\} \to \{1,2,3,4,5\}$,
means: $5^4=625$ and then reduced the number of injective functions that do exist $\{1,2\} \subseteq f[\{1,2,3,4\}]: 2 \cdot 5= 50$ and then
I got:  $625-50 \neq 30$


But I think I have a mistake in my method
any help?

As Eparoh pointed out in the comments, neither of your calculations is correct.
The number of injective functions $f: \{1, 2, 3, 4\} \to \{1, 2, 3, 4, 5\}$ such that $\{1, 2, 3\} \subseteq f[\{1, 2, 3, 4\}]$:  The statement means that the set $\{1, 2, 3\}$ is in the range of the injective function $f$.  There are four ways to select which element of the domain maps to $1$, three ways to select which of the remaining elements in the domain maps to $2$, and two ways to select which of the remaining elements in the domain maps to $3$.  Since the function is injective, the remaining element in the domain must map to $4$ or $5$.  Hence, there are
$$4 \cdot 3 \cdot 2 \cdot 2 = 48$$
injective functions $f: \{1, 2, 3, 4\} \to \{1, 2, 3, 4, 5\}$ such that $\{1, 2, 3\} \subseteq f[\{1, 2, 3, 4\}]$.
The number of injective functions $f: \{1, 2, 3, 4\} \to \{1, 2, 3, 4, 5\}$ such that $\{1, 2\} \not\subseteq f[\{1, 2, 3, 4\}]$:  The statement means that the elements $1, 2$ cannot both appear in the range of the injective function.  Since the function is injective, each of the four elements in the domain must map to a different element in the codomain, so the range of $f$ must include four of the five elements in the codomain.  Therefore, we can find the number of injective functions $f: \{1, 2, 3, 4\} \to \{1, 2, 3, 4, 5\}$ such that $\{1, 2\} \not\subseteq f[\{1, 2, 3, 4\}]$ by finding the number of injective functions $f: \{1, 2, 3, 4\} \to \{1, 2, 3, 4, 5\}$ that exclude $1$ or exclude $2$.
The number of injective functions $f: \{1, 2, 3, 4\} \to \{1, 2, 3, 4, 5\}$ that exclude $1$ from the range:  There are four ways to select which element of the domain maps to $2$, three ways to select which element of the domain maps to $3$, two ways to select which element of the domain maps to $4$, and one way to select which element of the domain maps to $5$.  Hence, there are
$$4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24$$
injective functions $f: \{1, 2, 3, 4\} \to \{1, 2, 3, 4, 5\}$ which exclude $1$ from the range.
The number of injective functions $f: \{1, 2, 3, 4\} \to \{1, 2, 3, 4, 5\}$ that exclude $2$ from the range:  By symmetry, there are
$$4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24$$
injective functions $f: \{1, 2, 3, 4\} \to \{1, 2, 3, 4, 5\}$ which exclude $2$ from the range.
Since an injective function $f: \{1, 2, 3, 4\} \to \{1, 2, 3, 4, 5\}$ must have four elements in its range, it is not possible for both $1$ and $2$ to be excluded from the range.  Hence, the number of injective functions $f: \{1, 2, 3, 4\} \to \{1, 2, 3, 4, 5\}$ such that $\{1, 2\} \not\subseteq f[\{1, 2, 3, 4\}]$ is
$$2 \cdot 4! = 48$$
