Counting total Number of Functions

In summary, the conversation discusses the problem of counting the total number of strictly increasing and non-decreasing functions from set A to set B, where A={1,2,3,...,n} and B={1,2,3,...,m} with 1<=n<=m. The speaker presents a formula for the total number of strictly increasing functions and discusses the importance of the function being injective. The listener suggests verifying the formula with small values of n and m.
  • #1
22990atinesh
143
1

Homework Statement



Consider Sets A={1,2,3,...,n} and B={1,2,3,...,m} where 1<=n<=m. F:A->B

1) Total no. of strictly increasing function i.e if x<y then f(x)<f(y)
2) Total no. of non decreasing function i.e if x<y then f(x)<=f(y)

How can we count total number of functions here

2. The attempt at a solution

1st element in set A has [(m-1)-n+1] choices
2nd ... [(m-2)-n+1] choices
3rd ... [(m-3)-n+1] choices

and so on

So for 1) I'm getting (m-1)(m-2)(m-3)...(m-2n+1)
Is it the correct approach
 
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  • #2
Do you agree in problem 1 that F must be injective? I think you do, so think how this interplays with the strictly increasing condition.

What I'm trying to get at is, there is a more conceptual way to count those functions.
 
  • #3
verty said:
Do you agree in problem 1 that F must be injective? I think you do, so think how this interplays with the strictly increasing condition.

What I'm trying to get at is, there is a more conceptual way to count those functions.

Sorry for 1) It should be (m-1)(m-2)(m-3)...(m-n) and If the function is injective then ans for 1) should be (m-n)(m-n-1)(m-m-2)...(m-2n+1).
 
Last edited:
  • #4
22990atinesh said:
Sorry for 1) It should be (m-1)(m-2)(m-3)...(m-n) and If the function is injective then ans for 1) should be (m-n)(m-n-1)(m-m-2)...(m-2n+1).

Why not try to verify your formulas by making n and m very small?
 
  • #5
verty said:
Why not try to verify your formulas by making n and m very small?

Examples doesn't always works. If Logic is correct then example will definitely work. :smile:
 

Related to Counting total Number of Functions

1. How do you define a function?

A function is a mathematical relationship between two sets of elements, where each element in the first set (called the domain) is assigned to exactly one element in the second set (called the range). The function can be represented as a rule or equation that maps the elements from the domain to the corresponding elements in the range.

2. What is the total number of functions?

The total number of functions can be calculated by using the formula 2^n, where n is the number of elements in the domain. This means that for every element in the domain, there are 2 possible mappings to the elements in the range. For example, if the domain has 3 elements, then the total number of functions is 2^3 = 8.

3. Are all functions bijective?

No, not all functions are bijective. A bijective function is one where each element in the range is mapped to by exactly one element in the domain. In other words, every element in the range has a unique pre-image in the domain. However, there are many functions that are not bijective, such as exponential functions or trigonometric functions.

4. How does the number of functions change with different sized domains?

The number of functions increases exponentially with the size of the domain. For example, if the domain has 4 elements, then the total number of functions is 2^4 = 16. But if the domain has 10 elements, then the total number of functions is 2^10 = 1024. This shows that as the domain gets larger, the number of possible functions increases significantly.

5. Can you give an example of a function with an infinite domain?

Yes, a classic example of a function with an infinite domain is the natural logarithm function. Its domain is all positive real numbers, which is an infinite set. Another example is the sine function, which has an infinite domain of all real numbers.

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