Let A = {a, b, c} and B = {1, 2, 3, 4}. Then the
number of elements in the set
C = {f : A $\to$ B |
2 $\in$ f(A) and f is not one-one} is ______.
Answer (integer)
19
Solution
The desired functions will contain either one
element or two elements in its codomain of
which '2' always belongs to f(A).
<br><br><b>Case 1 :</b> When 2 is the image of all element of set A.
<br><br>Number of ways this is possible = 1
<br><br><b>Case 2 :</b> When one image is 2 and other one image is one of {1, 3, 4}.
<br><br>Number of ways we can choose one of {1, 3, 4} is = <sup>3</sup>C<sub>1</sub>.
<br><br>Now divide 3 elements {a, b, c} of set A into two parts.
<br>We can do this ${{3!} \over {2!1!}}$ ways.
<br><br>Now map one part of set A into the element 2 of set B and map other part of set A into one of {1, 3, 4} of set B.
<br>We can do that 2! ways.
<br><br>So number of functions in this case
<br>= <sup>3</sup>C<sub>1</sub> $\times$ ${{3!} \over {2!1!}}$ $\times$ 2! = 18
<br><br>$\therefore$ Total number of functions = 1 + 18 = 19
About this question
Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations
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