
If P(S) denotes the set of all subsets of a given set S, then the number of one to one function from the set $s=\left\{ 1,2,3 \right\}$to the set P(S) is
(a) 336
(b) 8
(c) 36
(d) 320
Answer
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Hint: First, before proceeding for this, we must know that the number of subsets of the set S which is given by P(S) with number of elements as n is given by ${{2}^{n}}$. Then, we must know that formula for calculating the number of one to one functions from the set S where P(S) is the number of all subsets from S is given by ${}^{8}{{P}_{3}}$ for given question. Then, by solving the above expression, we get the number of one to one functions.
Complete step-by-step answer:
In this question, we are supposed to find the number of one to one function from the set $s=\left\{ 1,2,3 \right\}$to the set P(S) where P(S) denotes the set of all subsets of a given set S.
So, before proceeding for this, we must know that the number of subsets of the set S which is given by P(S) with number of elements as n i given by:
${{2}^{n}}$
Now, we have 3 elements in the above set S which gives the value of n as 3.
So, by substituting the value of n as 3, we get the number of subsets as:
${{2}^{3}}=8$
So, we get the total number of subsets as 8 which is given by $n\left( P\left( S \right) \right)=8$.
Now, we must know that formula for calculating the number of one to one functions from the set S where P(S) is the number of all subsets from S is given by:
${}^{8}{{P}_{3}}$
So, by solving the above expression, we get the number of one to one functions as:
$\begin{align}
& \dfrac{8!}{\left( 8-3 \right)!}=\dfrac{8\times 7\times 6\times 5!}{5!} \\
& \Rightarrow 8\times 7\times 6 \\
& \Rightarrow 336 \\
\end{align}$
So, we get the number of one to one functions from the set given is 336.
Hence, option (a) is correct.
Note: Now, to solve these type of the questions we need to know some of the basic formula for the permutation as ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$. Moreover, to find the factorial of the number n, multiply the number n with (n-1) till it reaches 1. To understand let us find the factorial of 4.
$\begin{align}
& 4!=4\times 3\times 2\times 1 \\
& \Rightarrow 24 \\
\end{align}$
Complete step-by-step answer:
In this question, we are supposed to find the number of one to one function from the set $s=\left\{ 1,2,3 \right\}$to the set P(S) where P(S) denotes the set of all subsets of a given set S.
So, before proceeding for this, we must know that the number of subsets of the set S which is given by P(S) with number of elements as n i given by:
${{2}^{n}}$
Now, we have 3 elements in the above set S which gives the value of n as 3.
So, by substituting the value of n as 3, we get the number of subsets as:
${{2}^{3}}=8$
So, we get the total number of subsets as 8 which is given by $n\left( P\left( S \right) \right)=8$.
Now, we must know that formula for calculating the number of one to one functions from the set S where P(S) is the number of all subsets from S is given by:
${}^{8}{{P}_{3}}$
So, by solving the above expression, we get the number of one to one functions as:
$\begin{align}
& \dfrac{8!}{\left( 8-3 \right)!}=\dfrac{8\times 7\times 6\times 5!}{5!} \\
& \Rightarrow 8\times 7\times 6 \\
& \Rightarrow 336 \\
\end{align}$
So, we get the number of one to one functions from the set given is 336.
Hence, option (a) is correct.
Note: Now, to solve these type of the questions we need to know some of the basic formula for the permutation as ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$. Moreover, to find the factorial of the number n, multiply the number n with (n-1) till it reaches 1. To understand let us find the factorial of 4.
$\begin{align}
& 4!=4\times 3\times 2\times 1 \\
& \Rightarrow 24 \\
\end{align}$
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