Discrete maths functions

Question

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

Comments

Please tell me where I went wrong. what i am getting by doing this?? please anyone.

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You applied inclusion wrong. 36-9+1. For two elements you will subtract ,for 3 elements you add and so on.Otherwise you are doing extra count

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ohkkk. .  Silly mistakes :(

Thanku  rahul sharma 5

Answer 1

Number of such injective function = Total injective function  - [which Pth element of set A match to Pth elemet of B]

Total injective function = ⁵p₃= 5*4*3= 60

Now, which Pth element of set A match to Pth elemet of B

Apply inclusion exclusion formula = (n(1)+n(2)+n(3) -n(1,2)-n(1,3) - n(2,3) + n(1,2,3)) = 3*n(1) -3*(1,2) +1*n(1,2,3)

When all 3 map to pth element;  Only 1 function

When any Two map to Pth element : ³c₂* ³p₁ = 9 function

When only 1 map to Pth element : ³c₁ * ⁴p₂ = 36 function

So,Number of such injective function = Total injective function  - [which Pth element of set A match to Pth elemet of B]

                                                                    = 60 - [36 -9+1]

                                                                    = 32

Comments

ok yes tnk u:)

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@Anu007 Sir , Can you plz explain a little bit how these ³p₁ and ⁴p₂ for injective functions are used ?

Shouldn't they be   ³p₂ and ⁴p₁ instead because at if taken two at a time in case 2 , Set A  will b having 2 elements in  and  1 element in set A in case 3 respectively ?beside that there only be 3 positions i.e. 1th,2nd and 3rd in set B where they can be mapped  then why did you take ⁴p_{2 not} ³p₂ ..Kindly help sir ..thank u

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plz explain how to map1 and map2 and map3