Question

How many distinct relations are possible from Set A $\rightarrow$ Set B?

Answer 1

Concept:
Let \[ A=\{G_1,G_2,G_3,G_4\} \] and \[ B=\{B_1,B_2,B_3\}. \] Thus, \[ n(A)=4,\qquad n(B)=3. \] The concepts used are:

• The number of relations from a set \(A\) to a set \(B\) is \(2^{n(A)\times n(B)}\).
• A function is injective (one-one) if different elements of the domain have different images.
• An equivalence relation must be reflexive, symmetric and transitive.
• A function is bijective if it is both one-one and onto.
  
  Here, \[ n(A)=4,\qquad n(B)=3. \] Hence, \[ n(A\times B)=4\times3=12. \] The number of possible relations is \[ 2^{12}=4096. \] Therefore, \[ \boxed{4096} \] relations are possible.

Answer 2

A function must assign exactly one image to every element of the domain. Here, \[ G_1\rightarrow B_1 \] and also \[ G_1\rightarrow B_2. \] Thus, the element \(G_1\) has two images. Hence, \(R\) is not even a function. Therefore, it cannot be an injective function. \[ \boxed{\text{No, }R\text{ is not an injective function because }G_1\text{ has two images.}} \]

Answer 3

To verify this, we check the three properties. 1. Reflexive: Every student belongs to the same colony as himself/herself. Therefore, \[ (x,x)\in R \] for every \(x\in A\). Hence, \(R\) is reflexive.
2. Symmetric: If student \(x\) and student \(y\) belong to the same colony, then student \(y\) and student \(x\) also belong to the same colony. Thus, \[ (x,y)\in R\Rightarrow(y,x)\in R. \] Hence, \(R\) is symmetric. 
3. Transitive: If \[ (x,y)\in R \] and \[ (y,z)\in R, \] then \(x,y,\) and \(z\) all belong to the same colony. Therefore, \[ (x,z)\in R. \] Hence, \(R\) is transitive. Since \(R\) is reflexive, symmetric and transitive, \[ \boxed{R\text{ is an equivalence relation}.} \] 

Answer 4

Here, \[ n(B)=3,\qquad n(A)=4. \] A bijective function must be both one-one and onto. Since the codomain has more elements than the domain, at least one element of \(A\) will remain without a pre-image. Therefore, no function from \(B\) to \(A\) can be onto. Hence, no function from \(B\) to \(A\) can be bijective. \[ \boxed{\text{No function }f:B\rightarrow A\text{ is bijective because }|B|<|A|.} \]