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In this article, we are discussing how to find number of functions from one set to another. For understanding the basics of functions, you can refer this: Classes (Injective, surjective, Bijective) of Functions. 

Nội dung chính

    How many onto functions are there from a set with 6 elements to a set with 4 elements?
    How many total functions are there from a set with three elements to a set with four elements?
    How many functions are there from a set with 3 elements to a set with 5 elements?
    How many functions are there from a set with two elements to a set with three elements?

Number of functions from one set to another: Let X and Y are two sets having m and n elements respectively. In a function from X

to Y, every element of X must be mapped to an element of Y. Therefore, each element of X has ‘n’ elements to be chosen from. Therefore, total number of functions will be n×n×n.. m times = nm. 
For example: X = a, b, c and Y = 4, 5. A function from X to Y can be represented in Figure 1. 
 

Considering all possibilities of mapping elements of X to

elements of Y, the set of functions can be represented in Table 1. 

Examples: Let us discuss gate questions based on this: 
 

    Q1. Let X, Y, Z be sets of sizes x, y and z respectively. Let W = X x Y. Let E be the set of all subsets of W. The number of functions from Z to E is: 
    (A) z2xy 
    (B) z x 2xy 
    (C)

    z2x + y 
    (D) 2xyz 

    Solution: As W = X x Y is given, number of elements in W is xy. As E is the set of all subsets of W, number of elements in E is 2xy. The number of functions from Z (set of z elements) to E (set of 2xy elements) is 2xyz. So the correct option is (D) 
     

    Q2. Let S denote the set of all functions f: 0,14 → 0,1. Denote by N the number of

    functions from S to the set 0,1. The value of Log2Log2N is ______. 
    (A) 12 
    (B) 13 
    (C) 15 
    (D) 16 

    Solution: As given in the question, S denotes the set of all functions f: 0, 14 → 0, 1. The number of functions from 0,14 (16 elements) to 0, 1 (2 elements) are 216. Therefore, S has 216 elements. Also, given, N denotes the number of function from S(216 elements) to 0, 1(2 elements).

    Therefore, N has 2216 elements. Calculating required value, 

    Log2(Log2 (2216)) =Log216 = 16 

    Therefore, correct option is (D). 
     

Number of onto functions from one set to another – In onto function from X to Y, all the elements of Y must be used. In the example of functions from X = a, b, c to Y = 4, 5, F1 and F2 given in Table 1 are not onto. In F1, element 5 of set Y is unused and element 4

is unused in function F2. So, total numbers of onto functions from X to Y are 6 (F3 to F8). 
 

    If X has m elements and Y has 2 elements, the number of onto functions will be 2m-2. 

    Explanation: From a set of m elements to a set of 2 elements, the total number of functions is 2m. Out of these functions, 2 functions are not onto (If all elements are mapped to 1st element of Y or all elements are mapped to

    2nd element of Y). So, number of onto functions is 2m-2.

    If X has m elements and Y has n elements, the number if onto functions are, 

     

Important notes

– 
 

    The formula works only if m ≥ n.
    If m < n, the number of onto functions is 0 as it is not possible to use all elements of Y.

Q3. The number of onto functions (surjective functions) from set X = 1, 2, 3, 4 to set Y = a, b, c is: 
(A) 36 
(B) 64 
(C) 81 
(D) 72 

Solution: Using m = 4 and n = 3, the number of onto functions is: 
34 – 3C1(2)4

+ 3C214 = 36. 

$begingroup$

Consider functions from a set with $5$ elements to a set with $3$ elements.
(a) How many functions are there?
(b) How many are one-to-one?
(c) How many are onto?

a) Each element mapped to $3$ images.
$3 cdot 3 cdot 3 cdot 3 cdot 3$

b) $0$

c) How do I

do this?

Edit: I tried doing this way.

EDIT: There can be a set of cardinality 3,1,1 or 2,2,1.

For 3,1,1: 5C3 * 2C1 * 1C1 * 3!

For 2,2,1: 5C2 * 3C2 * 1C1 * 3!

And i realized my 3! is wrong. Should be * 3 only. Why is that so?

asked Apr 28, 2022 8:47

RStyleRStyle

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5

$begingroup$

You correctly found that there are $3^5$ functions from a set

with five elements to a set with three elements. However, this counts functions with fewer than three elements in the range. We must exclude those functions. To do so, we can use the Inclusion-Exclusion Principle.

There are $binom31$ ways of excluding one element in the codomain from the range and $2^5$ functions from a set with five elements to the remaining two elements

in the codomain.

There are $binom32$ ways of excluding two elements in the codomain from the range and $1^5$ functions from a set with five elements to the remaining element in the codomain.

By the Inclusion-Exclusion Principle, the number of surjective (onto) functions from a set with five elements to a set with three elements is

$$3^5 – binom312^5 + binom321^5$$

answered Apr 28,

2022 9:11

N. F. TaussigN. F.

Taussig

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2

$begingroup$

Hint on c)

The “onto”-function will induce a partition of its domain (as any function) and this partition (actually the fibres of the function) will – because it is onto – have exactly $3$ elements. So to be found is in the first place how many such partitions exist. A fixed partition gives room for $3times2times1=6$ functions.

So you end up with: $$6timestextnumber of partitions on 1,2,3,4,5text

that have exactly 3text elements$$

Also have a look here (especially the counting of partitions).

A general formula for the number of onto-functions $1,dots,nto1,dots,k$ is: $$k!S(n,k)$$where $S(n,k)$ stands for the Stirling number of the second kind.

answered

Apr 28, 2022 8:55

drhabdrhab

143k9 gold badges70 silver badges191 bronze badges

$endgroup$

1

How many onto functions are there from a set with 6 elements to a set with 4 elements?

= 6 and S(4, 3) = 6. Thus, there are 36 onto functions.

How many total functions are there from a set with three elements to a set with four elements?

The correct choice is C. The total number of injective mappings from the set containing 3 elements into the set containing 4 elements is 4P3 = 4! = 24.

How many functions are there from a set with 3 elements to a set with 5 elements?

1 Answer. Image of each element of A can be taken in 3 ways. ∴ Number of functions from A to B = 35 = 243.

How many functions are there from a set with two elements to a set with three elements?

Thus there are 32 32 32 possible functions.

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