Let A = {0, 1, 2, 3, 4, 5, 6, 7}. Then the number of bijective functions f : A $\to$ A such that f(1) + f(2) = 3 $-$ f(3) is equal to
Answer (integer)
720
Solution
f(1) + f(2) = 3 $-$ f(3)<br><br>$\Rightarrow$ f(1) + f(2) = 3 + f(3) = 3<br><br>The only possibility is : 0 + 1 + 2 = 3<br><br>$\Rightarrow$ Elements 1, 2, 3 in the domain can be mapped with 0, 1, 2 only.<br><br>So number of bijective functions.<br><br>$\left| \!{\underline {\,
3 \,}} \right.$ $\times$ $\left| \!{\underline {\,
5 \,}} \right.$ = 720
About this question
Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Types of Functions
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