$A=\{1,2,3,4,5,6,7,8\} .$ How many functions $f: A \rightarrow A$ are there such that $f(1)<f(2)<f(3)$
[list=1]
[*] ${{8}\choose{3}}$


[*] ${{8}\choose{3}}5^{8}$


[*] ${{8}\choose{3}} 8^{5}$


[*] $\frac{8 !}{3 !} $
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1. We need to count the number of functions \( f: A \rightarrow A \) such that \( f(1) < f(2) < f(3) \). 
2. First, we choose 3 distinct values from the set \( A = \{1, 2, 3, 4, 5, 6, 7, 8\} \) for \( f(1), f(2), \) and \( f(3) \). The number of ways to choose 3 values out of 8 is given by the binomial coefficient:
   \[
   \binom{8}{3}
   \]
3. Once we have chosen 3 values, there is exactly one way to assign these values to \( f(1), f(2), \) and \( f(3) \) such that \( f(1) < f(2) < f(3) \).
4. For the remaining 5 elements \( f(4), f(5), f(6), f(7), \) and \( f(8) \), each can be any of the 8 elements in \( A \). Therefore, there are \( 8^5 \) ways to assign values to these 5 elements.
5. Combining these, the total number of functions \( f \) is:
   \[
   \binom{8}{3} \times 8^5
   \]

The final answer is \( \boxed{ \binom{8}{3} \times 8^5 } \)