Question

Let $f:A\to B$ be a real function, where $A=\{x_1,x_2,...,x_6\}$ and $B=\{y_1,y_2...,y_{10}\}$ given by $f\left(x\right)=y.$ Then the number of functions from $A$ to $B$ such that $f\left(x_1\right)<f\left(x_2\right)<f\left(x_3\right)<f\left(x_4\right)<f\left(x_5\right)<f\left(x_6\right)$ is

Answer 1

Selecting $6$ elements out of $10$ elements from $B$ can be done in ${}^{10}{C}_6$ ways.
Let these $6$ elements arranged in ascending order be $\{b_1,b_2,b_3,b_4,b_5,b_6\},b_i\in B$

All of these $6$ elements can be arranged in increasing order in such a way that $f\left(x_i\right)=b_i$ and $f\left(x_1\right)<f\left(x_2\right)<f\left(x_3\right)<f\left(x_4\right)<f\left(x_5\right)<f\left(x_6\right)$ can be done in $1$ way .

So, total number of required functions $={}^{10}{C}_6$.