Question

Consider two sets A = {2, 3, 5, 7, 11, 13} and B = {1, 8, 27}. Let f be a function from A to B such that for every element b in B, there is at least one element a in A such that f(a) = b. Then, the total number of such functions f is

• A. 537
• B. 540
• C. 667
• D. 665

Answer 1

The Correct Option is B

To find the total number of functions \( f: A \to B \) where every element in \( B \) is mapped to by at least one element from \( A \), we are looking for surjective functions.

We are given:

• Set A has 6 elements.
• Set B has 3 elements.

We need to find the number of surjective functions from a set of 6 elements to a set of 3 elements.

The formula for the number of surjective functions from a set of \( m \) elements to a set of \( n \) elements is:

\[ n! \times \left\{ \!\! \begin{array}{c} m \\ n \end{array} \!\! \right\} \]

Here, \(\left\{ \!\! \begin{array}{c} m \\ n \end{array} \!\! \right\}\) is the Stirling number of the second kind, which counts the ways to partition a set of \( m \) elements into \( n \) non-empty subsets.

Let's apply this to our problem:

(i) Calculate \( 3! \):

\[ 3! = 6 \]

 

(ii) Find the Stirling number of the second kind \( \left\{ \!\! \begin{array}{c} 6 \\ 3 \end{array} \!\! \right\} \):

The value is \( \left\{ \!\! \begin{array}{c} 6 \\ 3 \end{array} \!\! \right\} = 90 \).

(iii) Calculate the total number of surjective functions:

\[ 6 \times 90 = 540 \]

 

Therefore, there are 540 surjective functions.