Key Concepts and Formulas

• Bijective Function: A function $f: A \rightarrow B$ is bijective if it is both injective (one-to-one) and surjective (onto). For finite sets, this implies $|A| = |B|$, and each element in $A$ maps to a distinct element in $B$, with every element in $B$ being an image.
• Permutations: The number of ways to arrange $k$ distinct items from a set of $n$ distinct items, where order matters, is given by ${}^{n}P_k = \frac{n!}{(n-k)!}$.
• Combinations: The number of ways to choose $k$ distinct items from a set of $n$ distinct items, where order does not matter, is given by ${}^{n}C_k = \frac{n!}{k!(n-k)!}$.
• Relationship between Permutations and Combinations: ${}^{n}C_k \times k! = {}^{n}P_k$.

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Step-by-Step Solution

Step 1: Determine the size of the domain and codomain.

The domain is $A = \{1, 3, 5, \ldots, 99\}$. This is an arithmetic progression with first term 1, last term 99, and common difference 2. The number of elements in $A$ is $|A| = \frac{99 - 1}{2} + 1 = \frac{98}{2} + 1 = 49 + 1 = 50$.

The codomain is $B = \{2, 4, 6, \ldots, 100\}$. This is an arithmetic progression with first term 2, last term 100, and common difference 2. The number of elements in $B$ is $|B| = \frac{100 - 2}{2} + 1 = \frac{98}{2} + 1 = 49 + 1 = 50$.

Since $|A| = |B| = 50$, a bijective function can exist.

Step 2: Analyze the restriction on the function values.

The given condition is $f(3) \geq f(9) \geq f(15) \geq f(21) \geq \ldots \geq f(99)$. Let $X = \{3, 9, 15, 21, \ldots, 99\}$. This is an arithmetic progression with first term 3 and common difference 6. To find the number of elements in $X$, let $a_k = 99$. Using the formula $a_k = a_1 + (k-1)d$, we have $99 = 3 + (k-1)6$. This gives $96 = (k-1)6$, so $16 = k-1$, which means $k = 17$. Thus, there are 17 elements in the domain that are involved in this restriction.

Since $f$ is a bijective function, it must be injective, meaning that distinct elements in the domain map to distinct elements in the codomain. The elements $3, 9, 15, \ldots, 99$ are all distinct. Therefore, their images $f(3), f(9), f(15), \ldots, f(99)$ must also be distinct. This means the non-strict inequality must be a strict inequality: $f(3) > f(9) > f(15) > f(21) > \ldots > f(99)$

Step 3: Determine the number of ways to assign values to the restricted domain elements.

We need to choose 17 distinct values from the codomain $B$ (which has 50 elements) to be the images of the 17 elements in $X$. The number of ways to choose these 17 values is given by the combination formula ${}^{50}C_{17}$.

Once these 17 values are chosen, there is only one unique way to assign them to $f(3), f(9), \ldots, f(99)$ such that the condition $f(3) > f(9) > \ldots > f(99)$ is satisfied. The largest chosen value must be assigned to $f(3)$, the second largest to $f(9)$, and so on, down to the smallest chosen value being assigned to $f(99)$. So, the number of ways to map the restricted elements is ${}^{50}C_{17} \times 1 = {}^{50}C_{17}$.

Step 4: Determine the number of ways to assign values to the remaining domain elements.

We have 50 elements in the domain and 50 elements in the codomain. We have already assigned images for 17 domain elements, using 17 distinct codomain elements. The number of remaining domain elements is $50 - 17 = 33$. The number of remaining codomain elements is $50 - 17 = 33$.

Since the function must be bijective, these 33 remaining domain elements must be mapped bijectively to the 33 remaining codomain elements. The number of ways to establish a bijection between two sets of 33 elements is $33!$. This is because for the first remaining domain element, there are 33 choices in the remaining codomain; for the second, 32 choices; and so on.

Step 5: Calculate the total number of bijective functions.

The total number of bijective functions is the product of the number of ways to map the restricted elements and the number of ways to map the remaining elements. Total number of functions = (Ways to map restricted elements) $\times$ (Ways to map remaining elements) Total number of functions = ${}^{50}C_{17} \times 33!$

Now, we use the definition of combinations: ${}^{50}C_{17} = \frac{50!}{17!(50-17)!} = \frac{50!}{17!33!}$

Substituting this back into the total number of functions: Total number of functions = $\frac{50!}{17!33!} \times 33!$ Total number of functions = $\frac{50!}{17!}$

This expression can be written in terms of permutations. Recall that ${}^{n}P_k = \frac{n!}{(n-k)!}$. Here, we have $\frac{50!}{17!}$. We can see that $n=50$ and $n-k=17$. Solving for $k$, we get $k = 50 - 17 = 33$. Therefore, $\frac{50!}{17!} = {}^{50}P_{33}$.

The number of bijective functions is ${}^{50}P_{33}$.

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Common Mistakes & Tips

• Interpreting the Inequality: Always remember that for a bijective function, distinct inputs must map to distinct outputs. This converts a non-strict inequality ($\geq$) into a strict inequality ($>$).
• Order of Selection for Restricted Elements: When elements are arranged in a strictly increasing or decreasing order, selecting the elements is a combination, but the assignment is fixed (only one way).
• Independence of Mappings: The mapping of the restricted elements is independent of the mapping of the remaining elements. Multiply the possibilities from each part.

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Summary

The problem requires finding the number of bijective functions from a set of 50 odd numbers to a set of 50 even numbers, with a specific ordering constraint on 17 of the domain elements. First, we identified the sizes of the domain and codomain. Then, we analyzed the restriction, recognizing that the bijective nature of the function implies a strict inequality for the 17 specified domain elements. We calculated the number of ways to choose 17 distinct values from the codomain for these restricted elements (which is ${}^{50}C_{17}$) and noted that there's only one way to assign them according to the strict order. Finally, we calculated the number of ways to map the remaining 33 domain elements to the remaining 33 codomain elements, which is $33!$. Multiplying these two results gives ${}^{50}C_{17} \times 33! = \frac{50!}{17!33!} \times 33! = \frac{50!}{17!} = {}^{50}P_{33}$.

The final answer is \boxed{{ }^{50} P_{33}}. This corresponds to option (B).