Question

Let $f:A\to B$ be a function, where $A=\{x_1,x_2,x_3...,x_6\}$ and $B=\{y_1,y_2,y_3...,y_{10}\}$ 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

• A. ${}^{10}{C}_5$
• B. ${}^{10}{C}_3$
• C. ${}^{10}{C}_4$
• D. ${}^{10}{P}_6$

Answer 1

The correct option is A ${}^{10}{C}_5$

Total $5$ distinct outputs can be selected from $10$ in ${}^{10}{C}_5$ ways.
Let these $5$ elements arranged in ascending order be $\{b_1,b_2,b_3,b_4,b_5\},b_i\in B$

All of these $5$ elements can be arranged such that $f\left(x_1\right)<f\left(x_2\right)<f\left(x_3\right)=f\left(x_4\right)<f\left(x_5\right)<f\left(x_6\right)$ in $1$ way.

So, Number of required functions $={}^{10}{C}_5$.