JEE MAIN - Mathematics (2022 - 28th June Evening Shift - No. 28)
Let S = {1, 2, 3, 4}. Then the number of elements in the set { f : S $$\times$$ S $$\to$$ S : f is onto and f (a, b) = f (b, a) $$\ge$$ a $$\forall$$ (a, b) $$\in$$ S $$\times$$ S } is ______________.
Answer
37
Explanation
There are 16 ordered pairs in $S \times S$. We write all these ordered pairs in 4 sets as follows.
$A=\{(1,1)\}$
$B=\{(1,4),(2,4),(3,4)(4,4),(4,3),(4,2),(4,1)\}$
$C=\{(1,3),(2,3),(3,3),(3,2),(3,1)\}$
$D=\{(1,2),(2,2),(2,1)\}$
All elements of set $B$ have image 4 and only element of $A$ has image 1.
All elements of set $C$ have image 3 or 4 and all elements of set $D$ have image 2 or 3 or 4 .
We will solve this question in two cases.
Case I: When no element of set $C$ has image 3.
Number of onto functions $=2$ (when elements of set $D$ have images 2 or 3$)$
Case II: When atleast one element of set $C$ has image 3.
Number of onto functions $=\left(2^{3}-1\right)(1+2+2)$
$$ =35 $$
Total number of functions $=37$
$A=\{(1,1)\}$
$B=\{(1,4),(2,4),(3,4)(4,4),(4,3),(4,2),(4,1)\}$
$C=\{(1,3),(2,3),(3,3),(3,2),(3,1)\}$
$D=\{(1,2),(2,2),(2,1)\}$
All elements of set $B$ have image 4 and only element of $A$ has image 1.
All elements of set $C$ have image 3 or 4 and all elements of set $D$ have image 2 or 3 or 4 .
We will solve this question in two cases.
Case I: When no element of set $C$ has image 3.
Number of onto functions $=2$ (when elements of set $D$ have images 2 or 3$)$
Case II: When atleast one element of set $C$ has image 3.
Number of onto functions $=\left(2^{3}-1\right)(1+2+2)$
$$ =35 $$
Total number of functions $=37$
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