How many one to one functions are there from a set with five elements to sets with four elements?

How many subsets can be made from a set with 6 elements?

If a set has six elements, for example {A, B, C, D, E, F}, then it may have the following subsets: - the set itself - 6 sets of five elements - 15 sets of four elements - 20 sets of three elements - 15 sets of two elements - 6 sets of one element - 1 set with no elements (the null set), for a total of 64 sets, which is 2^6, or 2 to the 6th power.

Video Transcript

it's Claris when you right here. So for part A, the function is 1 to 1 if each element has a unique image. So since there are five elements in the domain and four elements in the range, it's not possible for the function to be one toe. Soldiers 0121 functions for part B. Okay, for part B of function is went to one if each element has a unique image. So there's five possible images since the scent got contains a range has five elements. So the first element there's five ways. The second element is four ways. The third is three ways. The fourth element is two ways, and the fifth element is one way. So we used the product will for five factorial me get 120 121 functions. I'll put a X for functions, rip apart. See their six possible images. So the first element has six ways. The second element has five ways. The third element has four ways. The fourth element has two ways. In the Fifth Element House, the fourth element has three ways in the Fifth Element has two ways. That's six times, five times four times three times two with just 721 to 100 shoes for D. There, seven possible images since the set that contains a Range House. Six elements. So the first element has seven ways. The second element has six ways. The third element has five ways to fourth. Element has six four ways. In the fifth Element has dream lease. So when we use the product, will you get seven times? Six times five times for times three, which is 2521 to 1 functions.

Video Transcript

in this question. It is us that We have two cents. Okay. And the function is defined between these two that said first is having for a limited. Okay. And the 2nd 1 is having five elements. Okay, What are the number of 1? one functions? Okay, so you can see that while making the valuable function in this first that will have five different options. This first element Second will have four grant options. And then third will have three different options. And the last one will have two different options. Okay, so we can say the number of 11 functions I went to four into three into And that is 1 20. Okay. In the second portion it is told that the number of elements, four elements in the first set. In the second set we have three limits. Okay, you can stay here that 11 function. Is it not possible because there are more elements in a and we cannot make the different images for all different elements. Okay, Similarly in the 3rd question We have four and horror limits in both the sets. Okay. And due to this reason while making the images, you can see The first element will have four options. Then second, we'll have three and then the next one will have then the next one we have one. So after making these possibilities product we get here. When people has the answer. Okay, in the the option we are asked that there are whole and six elements Okay, Since the four elements and be healthy six elements we can see here. The first element will have six options. Second we'll have five. Third one will have four and the last one will have three. So make the product of these possibilities. We can see without here 316 as the answer. Okay, In the last one held Again two sets, okay. The first set is having and the arguments and their dad many moments. Okay. And the second set is having and elephants Okay, no, what happens here? We can see this statement will have different options after making the major forecast When I listen to the reduces the name -1 Then I -2 then I'm -3. Okay the water, The number of 1, 1 Functions is the product of possibilities and -1 minus two -3 and how many factors should be there. That is the number of elements that is okay, so you can stay here and that should be the I am affected of not um okay, what should we know them? Because what is the mistake here is that first element will have any different options because in the second set we have an element Okay here we have an option Then N -1 then -2 and -3. So on So and multiplied and -1 and -2 and -3 so on and this should be an factors so It can be within us and and -1 and minus two and -30 so on. The last one is and minus M lest one Okay, this is dancer here for this question. So you can see the first question was having 20 is not possible 24 43 60.

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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, 2016 at 8:47

RStyleRStyle

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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 $\binom{3}{1}$ 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 $\binom{3}{2}$ 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

$$3^5 - \binom{3}{1}2^5 + \binom{3}{2}1^5$$

answered Apr 28, 2016 at 9:11

N. F. TaussigN. F. Taussig

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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 $3\times2\times1=6$ functions.

So you end up with: $$6\times\text{number of partitions on }\{1,2,3,4,5\}\text{ that have exactly }3\text{ elements}$$

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


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

answered Apr 28, 2016 at 8:55

How many one to one functions are there from a set with five elements to sets with four elements?

drhabdrhab

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How many one to one functions are there from a set of five elements to sets of five elements?

(d) 2520 one-to-one functions.

How many one to one functions are there from a set with 5 elements to the set with 4 number of elements?

Therefore, there are one-to-one functions from the set with 5 elements to the set with 4 elements.

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

The number of one to one functions is N!, because the max mapping to Y is N.