How many functions are there from a set with 5 elements to a set with 7 elements?

Use the formula {{n}^{m}}-\left[ ^{n}{{C}{1}}{{\left( n-1 \right)}^{m}}{{-}^{n}}{{C}{2}}{{\left( n-2 \right)}^{m}}+\cdots {{\left( -1 \right)}^{n}}{{\ }^{n}}{{C}_{n}} \right], here ‘m’ is the cardinality of the domain and ‘n’ is the cardinality of the co-domain to calculate the total number of surjective functions.

Let ‘k’ be the total number of surjective functions. Substitute n=4 and m=7 into the above mentioned formula.

k=47−[4C1(4−1)7 −4C2(4−2)7+4C3(4−3)7+(−1) 4 4C4]=47−[4C1(3 )7−4C2(2)7+4C3(1)7+4C4 ]=47−[4C1(3)7− 4C2(2)7+4C3+4C4]…(1)\begin{aligned} k&={{4}^{7}}-\left[ ^{4}{{C}_{1}}{{\left( 4-1 \right)}^{7}}{{-}^{4}}{{C}_{2}}{{\left( 4-2 \right)}^{7}}{{+}^{4}}{{C}_{3}}{{\left( 4-3 \right)}^{7}}+{{\left( -1 \right)}^{4}}{{\ }^{4}}{{C}_{4}} \right] \\ &={{4}^{7}}-\left[ ^{4}{{C}_{1}}{{\left( 3 \right)}^{7}}{{-}^{4}}{{C}_{2}}{{\left( 2 \right)}^{7}}{{+}^{4}}{{C}_{3}}{{\left( 1 \right)}^{7}}{{+}^{4}}{{C}_{4}} \right] \\&={{4}^{7}}-\left[ ^{4}{{C}_{1}}{{\left( 3 \right)}^{7}}{{-}^{4}}{{C}_{2}}{{\left( 2 \right)}^{7}}{{+}^{4}}{{C}_{3}}{{+}^{4}}{{C}_{4}} \right]\ldots \left( 1 \right) \end{aligned}

Use the formula ^{n}{{C}{r}}=\frac{n!}{r!\left( n-r \right)!} to calculate the value of ^{4}{{C}{1}}, ^{4}{{C}{2}}, ^{4}{{C}{3}} and ^{4}{{C}_{4}}.

4C1=4!1!3!=4⋅3!1!3!=4\begin{aligned} ^{4}{{C}_{1}}&=\frac{4!}{1!3!} \\&=\frac{4\cdot 3!}{1!3!} \\&=4 \end{aligned}

4C2= 4!2!2!=4⋅3⋅2!2!2!=6\begin{aligned} ^{4}{{C}_{2}}&=\frac{4!}{2!2!} \\&=\frac{4\cdot 3\cdot 2!}{2!2!} \\&=6 \end{aligned}

4C3=4!3!1! =4⋅3!3!1!=4\begin{aligned} ^{4}{{C}_{3}}&=\frac{4!}{3!1!} \\&=\frac{4\cdot 3!}{3!1!} \\&=4 \end{aligned}

4C4=4!4!0!=1 \begin{aligned} ^{4}{{C}_{4}}&=\frac{4!}{4!0!} \\&=1 \end{aligned}

Substitute the calculated value into the equation (1) and solve to get the required answer.

k =47−[4⋅37−6⋅27+4+1]=47−[4⋅37−6⋅27+5]=47 −7985=16384−7985=8399 \begin{aligned} k&={{4}^{7}}-\left[ 4\cdot {{3}^{7}}-6\cdot {{2}^{7}}+4+1 \right] \\&={{4}^{7}}-\left[ 4\cdot {{3}^{7}}-6\cdot {{2}^{7}}+5 \right] \\&={{4}^{7}}-7985 \\&=16384-7985 \\&=8399 \end{aligned}

Thus, the required number of surjective functions is 8399.

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7 elements that are not 1-1 ? Explain your reasoning fully.

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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

drhabdrhab

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