functions

Question

The number of ways possible to form injective function from set A set B where |A| = 3 and |B| = 5 such that $p^{th}$ element of set A cannot match with $p^{th}$ element of set B are________.

Comments

if you have confusion post their solution

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@Mk Utkarsh So what should be the steps to get this one.

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@`JEET

almost most of the madeeasy questions already asked here, please check for duplicates before posting.

https://gateoverflow.in/227449/made-easy-test

https://gateoverflow.in/170818/made-easy-test-series

https://gateoverflow.in/170311/discrete-maths-functions

Answer 1

|A|=5 |B|=3

total no of one one function =5P3=60

{a1,a2,a3} ===> {b1,b2,b3,b4,b5}

pth element map to pth eleement in n ways

n=n(A)+n(B)+n(C)-n(A ∩B)-n(B∩C)-n(C∩A) +n(A∩B∩C)

where A= a1 mapped to b1

           B=a2 mapped to b2

          C=a3 mapped to b3

no of ways in which a1 is mapped to b1=4P2=12 [ a1 -> b1 left combination {a2,a3}--->{b2,b3,b4,b5} ]

similarly for a2 and a3=12

A∩B=  a1 and a2 mapped to b1 and b2 respectively so left possible combination {a3 to {b3,b4,b5)) i.e 3P1=3

similarly B∩C,C∩A=4

so n=12 + 12 + 12 - 3 - 3 - 3 + 1

=28

no of ways in which pth elemement not mapped to pth =  60-n=60-28=32

Comments

Abhisek Tiwari 4 nice solution but one doubt

see u r saying // no of ways in which a1 is mapped to b1=4P2=12 [ a1 -> b1 left combination {a2,a3}--->{b2,b3,b4,b5}

okk i chose b1 for a1  .... now for a2 ,a3 ---> b2,b3,b4,b5  why are u using 4P2 instead of 4C2 ??

4C2 i think because  4elements can choose any element to arrange with them

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@Deepanshu 

{a2,a3}----->{b2,b3,b4,b5}  

if u choose 4C2 it will give b2,b4(lets say) that implies b4,b2 is also same combination but see  

(a2,b2}{a3,b4} and {a2,b4}{a3,b2} are two different function i.e why we take permutation.