Let A = {x1,x2,x3......,x7} and B = { y1 ,y2 ,y3 } be two sets containing seven and three distinct elements respectively. Then the total number of functions f: A \rightarrow B that are onto, if there exist exactly three elements x in A such that f(x) = y2, is equal to :

 

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

 

Answers (1)

As we learnt in

Number of Onto function -

f:A\rightarrow B

Such\: that\: n\left ( A \right )= m

&                    n\left ( B\right )=n

m\geqslant n

Number of onto functions = \sum_{r=1}^{n}\left ( -1 \right )^{n-r}n_{C_{r}}r^{m}

-

 

 No. of onto functions.

Since if there exist exactly three elements x in A. Which give single image y2. If means four elements of A will give image of  y1 and y2. So that (24 - 2) of ^{7}C_{3}.

=(16-2)^{7}C_{3}=14. ^{7}C_{3}

Correct option is 3.

 


Option 1)

This is an incorrect option.

Option 2)

This is an incorrect option.

Option 3)

This is an incorrect option.

Option 4)

This is the correct option.

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