Question

Given two real sets $A=\left\{a_{1} , a_{2} , a_{3} \ldots a_{2 n}\right\}$ and $B\left\{b_{1} , b_{2} , \ldots b_{n}\right\}.$ If $f:A \rightarrow B$ is a function such that every element of $B$ has an inverse image and $f\left(a_{1}\right)\leq f\left(a_{2}\right)\leq f\left(a_{3}\right)\leq f\left(a_{4}\right)\ldots \leq f\left(a_{2 n}\right),$ then the number of such mappings are

• A. $^{2 n}C_{n}$
• B. $^{2 n}C_{n - 1}$
• C. $^{2 n - 1}C_{n - 1}$
• D. $^{2 n + 1}C_{n}$

Answer 1

Answer: (C)

There is no loss of generality in considering the order of $b^{'}s$ as $b_{1} < b_{2} < b_{3}\ldots < b_{n}$ also given
that $f\left(a_{1}\right)\leq f\left(a_{2}\right)\cdot \cdot \cdot \cdot \leq f\left(a_{2 n}\right).$
Now suppose number of preimages of every $b_{i}$ are $x_{i}$ in numbers.
Therefore $x_{1}+x_{2}+\ldots +x_{n}=2n$ where $1\leq x_{i}\leq n+1........\left(\right.i\left.\right)$
Number of solutions of $\left(\right.i\left.\right)$ is ${}^{2 n - 1}C_{n - 1}^{}$ or ${}^{2 n - 1}C_{n}^{}$