[Math] no. of Strictly monotonic function in $f:\{1,2,3,4\}\rightarrow \{5,6,7,8,9\}$

algebra-precalculus

Of all the mapping that can be defined from $f:\{1,2,3,4\}\rightarrow \{5,6,7,8,9\}$ a mapping is

randomly selected. The chance that the selected mapping is strictly monotonic, is

My Try: Total no. of function from $A$ to $B$ in $f:A\rightarrow B$, is $n^m$

where $A$ contain $m-$elements and $B$ contain $n-$ elements.

So Total no. of function is $5^4$.

Now Total cases for Strictly monotonic function

(Means function which is either Strictly Increasing Increasing function or Strictly Decreasing functions.)

Strictly Increasing function: Means $f(1)<f(2)<f(3)<f(4)$

where values of $f(i)\in \{5,6,7,8,9\}, \forall i=1,2,3,4$

Similarly

Strictly Decreasing function: Means $f(1)>f(2)>f(3)>f(4)$

where values of $f(i)\in \{5,6,7,8,9\}, \forall i=1,2,3,4$

Now I did not understand How we can count Strictly Increasing and Strictly Decreasing function.

Help Required

Thanks

Best Answer

To count strictly increasing functions, consider some subset $S \subset \{5,6,7,8,9\}$ with four elements. There is extactly one strictly increasing function that maps onto $S$, so the number of such functions is the same that the numbers of subsets with $4$ elements, that is, $\binom 54$. Same reasoning for strictly decreasing functions.

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