Question

1.The number of surjection from A={1,2,……n}, \[n \ge 2\]onto B={a,b} is
a) \[{}^n{P_2}\]
b) \[{2^n} - 2\]
c) \[{2^n} - 1\]
d) None of these

Answer 1

Hint: We will check the number of element in A and number of element in B. If the number of onto function A greater than function B then we will use the formula \[\sum\limits_{r = 1}^N {{{( - 1)}^{N - r}}} {}^N{C_r}{r^M}\]
to find the surjection. And if the number of onto function A is less than B then the surjection is zero. Surjection means onto mapping.







Complete step by step solution:Given that , A={1,2,……n}, \[n \ge 2\]
                        B={a,b}
Since n(A)=n
                   n(B)=2
We know that if number of onto function from A(M) to B(N) , where M and N are the number of element in A and B function respectively then
\[\left( \begin{array}{l}\sum\limits_{r = 1}^N {{{( - 1)}^{N - r}}} {}^N{C_r}{r^M},M \ge N\\0,M \le N\end{array} \right)\]
Given , \[n \ge 2\]
First condition satisfied,
M=n, N=2
=\[\sum\limits_{r = 1}^N {{{( - 1)}^{N - r}}} {}^N{C_r}{r^M}\]
=\[\sum\limits_{r = 1}^2 {{{( - 1)}^{2 - r}}} {}^2{C_r}{r^n}\]
=\[{( - 1)^{2 - 1}}{}^2{C_1}{1^n} + {( - 1)^{2 - 2}}{}^2{C_2}{2^n}\]
=\[ - 2 + {2^n}\]
=\[{2^n} - 2\]



Option ‘B’ is correct



Note: Student whenever use the formula \[\sum\limits_{r = 1}^N {{{( - 1)}^{N - r}}} {}^N{C_r}{r^M}\] they put M=N which mislead them to find solution. M is the number of element of function A and N is the number of element in function B.