Question

Given two real sets $A=\left\{a_1,a_2,a_3\dots a_{2n}\right\}$ and $B\left\{b_1,b_2,\dots b_n\right\}.$ If $f:A\to B$ is a function such that every element of $B$ has an inverse image and $f\left(a_1\right)\le f\left(a_2\right)\le f\left(a_3\right)\le f\left(a_4\right)\dots \le f\left(a_{2n}\right),$ then the number of such mappings are

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

Answer 1

Answer: (C)

There is no loss of generality in considering the order of $b^{{}^{\prime }}s$ as $b_1<b_2<b_3\dots <b_n$ also given
that $f\left(a_1\right)\le f\left(a_2\right)\cdot \cdot \cdot \cdot \le f\left(a_{2n}\right).$
Now suppose number of preimages of every $b_i$ are $x_i$ in numbers.
Therefore $x_1+x_2+\dots +x_n=2n$ where $1\le x_i\le n+\mathrm{1........}(i)$
Number of solutions of $(i)$ is ${}^{2n-1}{C}_{n-1}^{}$ or ${}^{2n-1}{C}_n^{}$