The number of bijective functions f : {1, 3, 5, 7, ..., 99} → {2, 4, 6, 8, ..., 100}, such that f(3)

Q: Wait, the problem says $f(3) \ge f(9) \dots$, but you turned it into strict inequality. Why?

That is the most critical observation in this problem! The key is the word 'bijective'. A bijective function is one-to-one. If $f(3) = f(9)$, it would violate the one-to-one property because two distinct inputs would map to the same output. Therefore, equality is impossible, and the constraint must be strictly decreasing.

Q: Why do we multiply by $33!$ at the end?

Because the remaining 33 elements in the domain have no constraints! Once we have satisfied the condition for the 17 constrained elements, the remaining 33 elements in the domain are free to map to the remaining 33 elements in the codomain in any of the $33!$ possible permutations.

Q: I got $\binom{50}{17}$, but the answer is $^{50}P_{33}$. Where did I go wrong?

You likely stopped after choosing the values for the constrained set. You calculated the number of ways to pick the values, but you forgot that the rest of the function must still be defined. Multiplying by $33!$ accounts for the mapping of the remaining elements, which simplifies the expression into the permutation form $^{50}P_{33}$.

Question

The number of bijective functions $f : \{1, 3, 5, 7, \dots, 99\} 	o \{2, 4, 6, 8, \dots, 100\}$, such that $f(3) \ge f(9) \ge f(15) \ge f(21) \ge \dots \ge f(99)$, is

Options:

(A) $^{50}P_{17}$

(B) $^{50}P_{33}$

(C) $33! 	imes 17!$

(D) $\frac{50!}{2}$

Sigma Percentile · Accepted Answer

### The Dance of the Bijective Function

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a combinatorics problem; we are choreographing a dance between two sets.

Imagine you are standing before two grand ballrooms. The left ballroom holds our domain, and the right holds our codomain. Our goal is to pair every guest in the left room with a unique partner in the right room, following a very specific set of rules.

### Phase 1: The Ballroom Setup

First, let us count our guests. The domain $D$ is the set of odd numbers $\{1, 3, 5, \dots, 99\}$.

Using the arithmetic progression formula, we see there are exactly $50$ elements. The codomain $C$ is the set of even numbers $\{2, 4, 6, \dots, 100\}$.

Counting these, we find another $50$ elements. Since the number of elements is equal, a bijection is indeed possible.

### Phase 2: The Constrained Subset

Now, the problem introduces a strict choreographer. It singles out a subset of the domain: $S = \{3, 9, 15, \dots, 99\}$.

This is an arithmetic progression where the first term $a = 3$ and the common difference $d = 6$. We need to know how many guests are in this specific group.

Using the $n$-th term formula $l = a + (n-1)d$, we set $99 = 3 + (n-1)6$. Solving this, we find $96 = 6(n-1)$, which gives us $n-1 = 16$, so $n = 17$. We have $17$ guests who must follow a very rigid rule.

### Phase 3: The Bijection Trap

The rule states $f(3) \ge f(9) \ge \dots \ge f(99)$. But wait! We must pause.

The problem tells us the function is **bijective**. A bijective function must be **injective**, meaning no two distinct inputs can share the same output.

If $f(3)$ were equal to $f(9)$, the function would fail to be injective. Therefore, the 'greater than or equal to' sign is a clever disguise. It must be a **strict inequality**: 
$$f(3) > f(9) > f(15) > \dots > f(99)$$

### Phase 4: The Selection and Arrangement

We have $17$ guests who must be mapped to $17$ distinct values from the codomain. First, we must choose $17$ values out of the $50$ available in the codomain.

The number of ways to do this is $\binom{50}{17}$.

Now, here is the beauty of the strict inequality. Once we have chosen these $17$ values, is there any choice in how we assign them? Absolutely not!

The largest value must go to $f(3)$, the second largest to $f(9)$, and so on, down to the smallest value for $f(99)$. There is exactly $1$ way to arrange them. So, our current count is $\binom{50}{17} 	imes 1$.

### Phase 5: The Remaining Guests

We are not done yet. We have used $17$ elements from the domain and $17$ from the codomain.

That leaves $50 - 17 = 33$ elements in both sets. These remaining guests have no constraints.

They can dance with each other in any way they please. The number of ways to map $33$ elements to $33$ elements is $33!$.

### Phase 6: The Grand Finale

To find the total number of bijective functions, we multiply our selection-and-arrangement count by the freedom of the remaining elements:

$$	ext{Total} = \binom{50}{17} 	imes 33!$$

Let us expand the combination:

$$\binom{50}{17} = \frac{50!}{17! 	imes 33!}$$

Substituting this back into our equation:

$$	ext{Total} = \frac{50!}{17! 	imes 33!} 	imes 33!$$

The $33!$ terms cancel out with elegant precision, leaving us with:

$$\frac{50!}{17!}$$

Recall the definition of a permutation: $^{n}P_{r} = \frac{n!}{(n-r)!}$. Here, $50 - 33 = 17$.

Thus, our result is exactly $^{50}P_{33}$. We have successfully navigated the constraints and arrived at the solution. Remember, in JEE Advanced, it is not just about the calculation; it is about understanding the constraints that define the system. You have done well.

Animated Solution for Mathematics - Permutations and Combinations: The number of bijective functions $f : {1, 3, 5, 7, \dots, 99} \to {2, 4, 6, 8, \dots, 100}$ , such that $f (3) \geq f (9) \geq f (15) \geq f (21) \geq \dots \geq f (99)$ , is

Select Answer:

Visualized Solution

Defining the Domain and Codomain

The Constrained Subset

Counting the Elements in A.P.

Strict Inequality from Bijection

Choosing Codomain Values

Mapping the Constrained Elements

Mapping the Remaining Elements

Total Bijective Functions

Final Answer

The Sigma Insight: Combinations and Selection

The Dance of the Bijective Function

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

Phase 6

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