# Let X be a set with exactly 5 elements and let Y be a set with exactly 7 elements. If $\alpha $

is the number of one-one functions from X to Y and $\beta $ is the number of onto functions from Y to X, then the value of $\dfrac{1}{51}\left( \alpha -\beta \right)$ is

Last updated date: 24th Mar 2023

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Answer

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Hint: Use the fact that if m be the number of elements in a set X and n be the number of elements in a set

Y, and if $n\ge m$, then the number of one-one functions from X to Y is given by the formula

$\dfrac{n!}{\left( n-m \right)!}$. Further use the fact that the total number of onto functions from a set X with m elements and another set Y with n elements, such that $m\ge n$ is given by the sum $\sum\limits_{k=0}^{n}{-{{1}^{k}}}\left( \begin{matrix} n \\ k \\

\end{matrix} \right){{\left( n-k \right)}^{m}}$. . These values of $\alpha $ and $\beta $ can then be used to calculate the required value of $\dfrac{1}{51}\left( \alpha -\beta \right)$.

Complete step by step solution:

For mapping functions from set X having 5 elements to set Y having 7 elements, these functions can be

either one-one or many-one. The total number of one-one functions can be calculated using the formula

$\dfrac{n!}{\left( n-m \right)!}$, where n is the number of elements in Y and m is the number of elements

in X.

Thus, for the given question, $m=5$ and $n=7$. Using these values in the formula, we get

$\begin{align}

& \alpha =\dfrac{7!}{\left( 7-5 \right)!} \\

& \Rightarrow \alpha =\dfrac{7!}{2!} \\

& \Rightarrow \alpha =7\times 6\times 5\times 4\times 3 \\

& \Rightarrow \alpha =2520 \\

\end{align}$

Thus, the required value of $\alpha $ is 2520.

For the calculation of $\beta $, consider the mapping of functions from Y to X. The total number of onto

functions from a set Y having m elements to another set X having n elements, where $m\ge n$ is given by

the formula $\sum\limits_{k=0}^{n}{-{{1}^{k}}}\left( \begin{matrix}

n \\

k \\

\end{matrix} \right){{\left( n-k \right)}^{m}}$.

Thus, we calculate this sum with \[m=7\] and \[n=5\] as

$\begin{align}

& \beta =\sum\limits_{k=0}^{5}{{{\left( -1 \right)}^{k}}}\left( \begin{matrix}

5 \\

k \\

\end{matrix} \right){{\left( 5-k \right)}^{7}} \\

& \Rightarrow \beta ={{\left( -1 \right)}^{0}}\left( \begin{matrix}

5 \\

0 \\

\end{matrix} \right){{\left( 5-0 \right)}^{7}}+{{\left( -1 \right)}^{1}}\left( \begin{matrix}

5 \\

1 \\

\end{matrix} \right){{\left( 5-1 \right)}^{7}}+{{\left( -1 \right)}^{2}}\left( \begin{matrix}

5 \\

2 \\

\end{matrix} \right){{\left( 5-2 \right)}^{7}}+{{\left( -1 \right)}^{3}}\left( \begin{matrix}

5 \\

3 \\

\end{matrix} \right){{\left( 5-3 \right)}^{7}} \\

& \ \ \ \ \ \ \ \ \ \ +{{\left( -1 \right)}^{4}}\left( \begin{matrix}

5 \\

4 \\

\end{matrix} \right){{\left( 5-4 \right)}^{7}}+{{\left( -1 \right)}^{5}}\left( \begin{matrix}

5 \\

5 \\

\end{matrix} \right){{\left( 5-5 \right)}^{7}} \\

& \Rightarrow \beta =1\times {{5}^{7}}-5\times {{4}^{7}}+10\times {{3}^{7}}-10\times

{{2}^{7}}+5\times {{1}^{7}} \\

& \Rightarrow \beta =5\left( {{5}^{6}}-{{4}^{7}} \right)+10\left( 2187-128 \right)+5 \\

& \Rightarrow \beta =5\left( 15625-16384 \right)+10\times 2059+5 \\

& \Rightarrow \beta =5\times \left( -759 \right)+20590+5 \\

& \Rightarrow \beta =20595-3795 \\

& \Rightarrow \beta =16800 \\

\end{align}$

Thus, the value of $\beta $ comes out to be 16800. This gives the value of $\dfrac{1}{51}\left( \beta -

\alpha \right)$ as

$\begin{align}

& \dfrac{1}{51}\left( \beta -\alpha \right)=\dfrac{1}{51}\left( 16800-2520 \right) \\

& \Rightarrow \dfrac{1}{51}\left( \beta -\alpha \right)=\dfrac{1}{51}\left( 14280 \right) \\

& \Rightarrow \dfrac{1}{51}\left( \beta -\alpha \right)=280 \\

\end{align}$

Thus the required value of $\dfrac{1}{51}\left( \beta -\alpha \right)$ is 280.

Note: The conditions for the calculation of one-one function and the calculation of the number of onto

functions are very important and to be kept in mind. These conditions, $n\ge m$ for one-one functions

and $m\ge n$ for onto functions is not only preliminary to the application of formulae but also necessary

for the existence of one-one and onto functions. If these conditions are violated, the number of one-one

functions and onto functions will both become 0 in their respective cases.

Y, and if $n\ge m$, then the number of one-one functions from X to Y is given by the formula

$\dfrac{n!}{\left( n-m \right)!}$. Further use the fact that the total number of onto functions from a set X with m elements and another set Y with n elements, such that $m\ge n$ is given by the sum $\sum\limits_{k=0}^{n}{-{{1}^{k}}}\left( \begin{matrix} n \\ k \\

\end{matrix} \right){{\left( n-k \right)}^{m}}$. . These values of $\alpha $ and $\beta $ can then be used to calculate the required value of $\dfrac{1}{51}\left( \alpha -\beta \right)$.

Complete step by step solution:

For mapping functions from set X having 5 elements to set Y having 7 elements, these functions can be

either one-one or many-one. The total number of one-one functions can be calculated using the formula

$\dfrac{n!}{\left( n-m \right)!}$, where n is the number of elements in Y and m is the number of elements

in X.

Thus, for the given question, $m=5$ and $n=7$. Using these values in the formula, we get

$\begin{align}

& \alpha =\dfrac{7!}{\left( 7-5 \right)!} \\

& \Rightarrow \alpha =\dfrac{7!}{2!} \\

& \Rightarrow \alpha =7\times 6\times 5\times 4\times 3 \\

& \Rightarrow \alpha =2520 \\

\end{align}$

Thus, the required value of $\alpha $ is 2520.

For the calculation of $\beta $, consider the mapping of functions from Y to X. The total number of onto

functions from a set Y having m elements to another set X having n elements, where $m\ge n$ is given by

the formula $\sum\limits_{k=0}^{n}{-{{1}^{k}}}\left( \begin{matrix}

n \\

k \\

\end{matrix} \right){{\left( n-k \right)}^{m}}$.

Thus, we calculate this sum with \[m=7\] and \[n=5\] as

$\begin{align}

& \beta =\sum\limits_{k=0}^{5}{{{\left( -1 \right)}^{k}}}\left( \begin{matrix}

5 \\

k \\

\end{matrix} \right){{\left( 5-k \right)}^{7}} \\

& \Rightarrow \beta ={{\left( -1 \right)}^{0}}\left( \begin{matrix}

5 \\

0 \\

\end{matrix} \right){{\left( 5-0 \right)}^{7}}+{{\left( -1 \right)}^{1}}\left( \begin{matrix}

5 \\

1 \\

\end{matrix} \right){{\left( 5-1 \right)}^{7}}+{{\left( -1 \right)}^{2}}\left( \begin{matrix}

5 \\

2 \\

\end{matrix} \right){{\left( 5-2 \right)}^{7}}+{{\left( -1 \right)}^{3}}\left( \begin{matrix}

5 \\

3 \\

\end{matrix} \right){{\left( 5-3 \right)}^{7}} \\

& \ \ \ \ \ \ \ \ \ \ +{{\left( -1 \right)}^{4}}\left( \begin{matrix}

5 \\

4 \\

\end{matrix} \right){{\left( 5-4 \right)}^{7}}+{{\left( -1 \right)}^{5}}\left( \begin{matrix}

5 \\

5 \\

\end{matrix} \right){{\left( 5-5 \right)}^{7}} \\

& \Rightarrow \beta =1\times {{5}^{7}}-5\times {{4}^{7}}+10\times {{3}^{7}}-10\times

{{2}^{7}}+5\times {{1}^{7}} \\

& \Rightarrow \beta =5\left( {{5}^{6}}-{{4}^{7}} \right)+10\left( 2187-128 \right)+5 \\

& \Rightarrow \beta =5\left( 15625-16384 \right)+10\times 2059+5 \\

& \Rightarrow \beta =5\times \left( -759 \right)+20590+5 \\

& \Rightarrow \beta =20595-3795 \\

& \Rightarrow \beta =16800 \\

\end{align}$

Thus, the value of $\beta $ comes out to be 16800. This gives the value of $\dfrac{1}{51}\left( \beta -

\alpha \right)$ as

$\begin{align}

& \dfrac{1}{51}\left( \beta -\alpha \right)=\dfrac{1}{51}\left( 16800-2520 \right) \\

& \Rightarrow \dfrac{1}{51}\left( \beta -\alpha \right)=\dfrac{1}{51}\left( 14280 \right) \\

& \Rightarrow \dfrac{1}{51}\left( \beta -\alpha \right)=280 \\

\end{align}$

Thus the required value of $\dfrac{1}{51}\left( \beta -\alpha \right)$ is 280.

Note: The conditions for the calculation of one-one function and the calculation of the number of onto

functions are very important and to be kept in mind. These conditions, $n\ge m$ for one-one functions

and $m\ge n$ for onto functions is not only preliminary to the application of formulae but also necessary

for the existence of one-one and onto functions. If these conditions are violated, the number of one-one

functions and onto functions will both become 0 in their respective cases.

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