Question

Let $A=\{x_1,x_2,x_3,\dots ,x_8\},B=\{y_1,y_2,y_3\}$ then the total number of functions from $A$ to $B$ such that all the elements of $B$ has atleast one pre image and there are exactly four elements in $A$ having image as $y_3,$ are

• A. $14\times {}^8{C}_4$
• B. $24\times {}^8{C}_4$
• C. $14\times {}^4{C}_4$
• D. $16\times {}^4{C}_4$

Answer 1

The correct option is A $14\times {}^8{C}_4$

Four elements from set $A$ can be selected in ${}^8{C}_4$ ways.
Now the remaining $4$ elements of $A$ can be connected with two elements of $B$ in $2^4-{}^2{C}_1(1{)}^4=14$ ways.
So, the required number of functions
$=14\times {}^8{C}_4$