Find the number of bijective function such that $f(3) \geq f(9) \geq \ldots \geq f(99)$

Question

For a function $f:\{1,3,5,7,\ldots,99\} \to \{2,4,6,8,\ldots,100\}$. Find the no of bijective functions such that $f(3) \geq f(9) \geq \ldots \geq f(99)$ is: The sequence has a gap of $6$. So, it is like $3,9,15,21,27,\ldots$ up to $ 99$.

• How I am trying to solve and understand the question

1. As per law of finding number of bijective functions:
   If $n(a) = n(b)$. Then, number of possible bijective functions $= m!$

Here, $n$ is a function name, $m$ is total number of elements in the sets $A$, $B$.

2. What does it mean? It means that for every $1$ element of set $A$, $1$ element of set $B$ is possible. Now, using formula what we mean in basic term is: $f(1) = 2,4,\ldots,100$ ($50$ ways). Now, $f(3)$ will be $= 4,6,8,\ldots,100$.

3. Now coming to the question: They say $f(3) \geq f(9) \geq \ldots \geq f(99)$. As I think this means: The value you find in $f(3)$ has to be greater or equal to the value I find in $f(9)$ but I don’t know the condition for $f(5)$, $f(7)$.

I am unable to think how to solve this to get that condition of $f(3)$ and neither getting what to do $f(5)$, $f(7)$.

Answer 1

Let $B = \{2,4,6,8,\ldots,100\}$, $n=|B|$, $k = |\{3,9,15, ... 99\}|$.

1. Choose a subset of $B$ of size $k$.
2. Assign values from that subset to $f(3), f(9), f(15), ... f(99)$, such that condition $f(3) > f(9) > ... > f(99)$ holds.
3. Take the remaining elements of $B$ (there are $n-k$ of them) and assign them to $f(1), f(5), f(7), f(11),... f(97)$ in any order.

With different choices in steps 1, 2, 3 this algorithm never produces the same bijection. Also this algorithm can produce any bijection (prove it).

How many ways to do step (1)? How many ways ways to do step (2)? How many ways to do step (3)? Multiply these three numbers, and that will be the number of bijections.