Question

If n(A)≥n(B), and n(B) = 2 , then how many onto functions are possible ?​

Answer 1

$\large\underline{\sf{Solution-}}$

It is given that

• n(A) ≥ n(B)

and

• n(B) = 2

Let assume that

• Number of elements in set A be 'n'.

Now,

• Set B has 2 elements.

Such that n ≥ 2

We know,

If A and B are two non - empty sets such that n(A) = n and n(B) = m, such that n(A) ≥ n(B), then every element of B has pre - image in set A, then number of onto functions from A to B is given as

$\rm :\longmapsto\:Number \: of \: onto \: functions = \displaystyle\sum_{r=1}^m \: {( - 1)}^{m - r} \: ^mC_r \: {r}^{n}$

According to statement,

We have,

• n(A) = n

• n(B) = 2

• Such that n(A) ≥ n(B)

So,

$\rm :\longmapsto\:Number \: of \: onto \: functions = \displaystyle\sum_{r=1}^2 \: {( - 1)}^{2 - r} \: ^2C_r \: {r}^{n}$

$\rm \: \: = \: {( - 1)}^{2 - 1} \: ^2C_1 \: {1}^{n} + {( - 1)}^{2 - 2} \: ^2C_2 \: {2}^{n}$

$\rm \: \: = \: {( - 1)}^{1} \: \times 2 \: \times 1 + {( - 1)}^{0} \: \times 1 \times \: {2}^{n}$

$\rm \: \: = \: {2}^{n} - 2$

Hence,

$\red{\bf :\longmapsto\:Number \: of \: onto \: functions \: = \: {2}^{n} - 2}$

Additional Information :-

If A and B are two non - empty sets such that n(A) = n and n(B) = m, such that n(A) < n(B), then every element of B has pre - image in set A, then number of onto functions from A to B is given as

$\rm :\longmapsto\:Number \: of \: onto \: functions \: = \: 0$

Number of one - one functions

If A and B are two non - empty sets such that n(A) = n and n(B) = m, then every element of A has unique - image in set B, then number of one-one functions from A to B is given as

$\begin{gathered}\begin{gathered}\bf\: Number \: of \: one \: one \: functions-\begin{cases} &\sf{0 \: \: if \: n > m} \\ &\sf{^mP_n \: if \: n \leqslant m} \end{cases}\end{gathered}\end{gathered}$