The number of functions f from {1, 2, 3, ...., 20} onto {1, 2, 3, ....., 20} such that f(k) is a mul

Q: Why do we assume it's a bijection? The question only says 'onto'.

In finite sets, 'onto' (surjective) implies 'one-to-one' (injective). If you have 20 items and 20 slots, and every slot must be filled, you cannot have two items in one slot. It's a perfect one-to-one correspondence!

Q: Why is it $6!$ and not just $^6C_5$?

Great question! $^6C_5$ only *selects* the 5 targets. But the function cares about *which* input maps to *which* target. Since the inputs are distinct (4, 8, 12, 16, 20), the order matters. That's why we use permutations: $^6P_5 = 6!$.

Q: Can I map a multiple of 4 to a non-multiple of 3?

No, the problem explicitly states: 'whenever $k$ is a multiple of 4, $f(k)$ is a multiple of 3'. This is a hard constraint. You must satisfy this condition first before dealing with the rest of the elements.

Question

The number of functions $f$ from $\{1, 2, 3, \dots, 20\}$ onto $\{1, 2, 3, \dots, 20\}$ such that $f(k)$ is a multiple of 3, whenever $k$ is a multiple of 4, is:

Options:

(A) $5^6 	imes 15$

(B) $6^5 	imes (15)!$

(C) $(15)! 	imes 6!$

(D) $5! 	imes 6!$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with finding the number of bijective functions $f: \{1, 2, \dots, 20\} 	o \{1, 2, \dots, 20\}$ such that for every $k$ which is a multiple of $4$, $f(k)$ is a multiple of $3$.

In a finite set of equal size, an 'onto' function is necessarily a bijection. This means we are looking for the number of permutations of the set $\{1, 2, \dots, 20\}$ that satisfy the given condition.

### The Special Forces

First, let us identify the inputs that are multiples of $4$. These are the elements of the set $S_4 = \{4, 8, 12, 16, 20\}$. There are exactly $5$ such elements.

Next, we identify the available targets in the codomain that are multiples of $3$. These are the elements of the set $S_3 = \{3, 6, 9, 12, 15, 18\}$. There are exactly $6$ such elements.

Since the function must be a bijection, each of the $5$ inputs in $S_4$ must be mapped to a unique target in $S_3$. The number of ways to choose $5$ targets out of $6$ and assign them to the $5$ inputs is given by the permutation formula:

$$^6P_5 = \frac{6!}{(6-5)!} = 6! = 720$$

### The Remaining Army

After mapping the $5$ special inputs, we have used $5$ inputs from the domain and $5$ targets from the codomain. This leaves us with $20 - 5 = 15$ remaining inputs and $20 - 5 = 15$ remaining targets.

Because the function must be a bijection, these remaining $15$ inputs must be mapped to the remaining $15$ targets. Since there are no further restrictions on these elements, they can be paired in any way possible.

The number of ways to arrange these $15$ items into $15$ slots is simply $15!$.

### The Grand Finale

By the **Fundamental Principle of Counting**, we multiply the number of ways to perform these two independent steps to find the total number of valid functions.

The total number of such functions is:

$$N = 6! 	imes 15!$$

This result represents the total count of bijective mappings satisfying the given constraints, demonstrating how breaking a complex problem into smaller, manageable pieces leads to a clear and elegant solution.

Animated Solution for Mathematics - Permutations and Combinations: The number of functions $f$ from ${1, 2, 3, \dots, 20}$ onto ${1, 2, 3, \dots, 20}$ such that $f (k)$ is a multiple of 3, whenever $k$ is a multiple of 4, is:

Select Answer:

Visualized Solution

Defining Domain and Codomain

The Bijective Requirement

Constraint: Multiples of 4

Target: Multiples of 3

Mapping the Special Elements

Mapping the Remaining Elements

Total Number of Functions

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Special Forces

The Remaining Army

The Grand Finale

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