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The number of ways possible to form injective function from set A to 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 _________.

My Attempt:

The solution to this one will be like below

Total number of injective functions from 3 element set to 5 element set - Total number of injective functions in which either the first, or second or third element from set A matches with their respective element in Set B.

So, Total number of injective functions from a 3 element set to a 5 element set = $5P_3=60$

Now Case 1: Element $i$ matches with $i^{th}$ element of set B- Number of ways to select this $i^{th}$ element=$3C_1$ and now number of injective functions from 2 element set to 4 element set=$4P_2$. So total ways=$3C_1*4P_2=36$

Case 2: Element $i$ and $j$ match with $i^{th}$ and $j^{th}$ element of the set B.-Number of ways to select $i,j=3C_2$ and now number of injective functions from a 1 element set to a 3 element set = $3P_1$. So total ways here=9.

Case 3: Element $i,j,k$ match respectively with their elements in Set B.-Only 1 way.

So, total number of injective functions from a 3 element set to a 5 element set in which $p^{th}$ element of set A matches with $p^{th}$ element of set B=36-9+1 (By inclusion-exclusion principle)=28

So, my final answer should be 60-28=32

But they have given the solution like this below

I think in case 2, it should be like $3C_2*3P_1$ instead of $4P_1$ because after $i^{th}$ and $j^{th}$ elements are mapped to their respective elements, now only 1 element from Set A need to be mapped to remaining 3 elements in Set B, considering that function has to be injective, so total ways must be 3.

What should be the correct way?

in Calculus
recategorized | 161 views
0

Ayush Upadhyaya, yes your way is right

but i have doubt that, cases are not overlapped?

i mean, n(A U B U C) = n(A)+n(B)+n(C) - n(A ∩ B) - n(B ∩ C) - n( A ∩ C) + n(A  ∩ B ∩ C )

A---> case 1 :- i is matched with i

B---> case 2 :- i is matched with i and j is matched with j

C---> case 3 :- i is matched with i and j is matched with j and k is matched with k

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@Shaik-No they are. Case 1 makes sure element i surely matches with i and all other can match with anyone.

Case 2 makes sure i and j match with i and j and rest with anyone. So, there might be a case in case 1 where i matches with i and among the rest j also matches with j.
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in case 1:-

let 1 is matched with 1===> in all possible cases, just take in any case 2 is matched with 2

in case 2:-

let 1 is matched with 1 and 2 is matched with 2 ===> didn't overlap with case1 ?
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This case will come under case 1 and case 2 both

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That's what i am saying also...

there are some possibles counted more than once...

therefore you have to use this formula n(A U B U C) = n(A)+n(B)+n(C) - n(A ∩ B) - n(B ∩ C) - n( A ∩ C) + n(A  ∩ B ∩ C )

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yes it should be 32.
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But i am getting 36+9+1-9-1-1+1 ( terms with respect to formula ) = 36

Required =  60-36 = 24. ... Correct me if i am wrong
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YES i am getting ans 32 too. I guess it is correct!!
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can you share your answer?

## 1 Answer

+2 votes
Best answer

We can solve this problem easily by using Inclusion-Exclusion principle.

First we will count the number of onto functions where at least one element of $A$ maps to the corresponding element of $B.$

$|1 \cup 2 \cup 3|= |1| + |2| + |3| - |1 ∩ 2| - |1∩3| - |2∩ 3| + |1∩ 2 ∩ 3|$
where, $|x|$ represents no. of possible functions in which $x^{th}$ element of $A$ is mapped to $x^{th}$ element of $B.$

Lets consider the case where first element from set $A$ is mapped to first element of set $B.$

First element of $A$ can be mapped to first element of $B$ in only $1$ way.

Now, in calculating the total number of functions where this happens, $2^{nd}$ and $3^{rd}$ elements can be mapped to in $4\times 3$ ways and not $4\times 4$ as we are forming one-one functions and not general functions.

So, total no. of functions in which $1^{st}$ element of $A$ is mapped to the first element of $B$ $= 1\times 4\times 3=12.$

Similarly for $|2|$ and $|3|$ also we will get $12$ ways. So, $|1| + |2| + |3|= 36.$

Now,  $|1 ∩ 2| =$ number of one-one functions where $1$ and $2$ are mapped to first and second elements respectively.

$=1\times 1 \times 3.$

Similarly, $|1 ∩ 2| = |1 ∩ 2|=3$ and $|1∩ 2 ∩ 3|=1.$

So, $|1 \cup 2 \cup 3 |=36-9+1=28.$

Total no. of one-one functions possible from $A$ to $B=5\times 4\times 3=60.$

Hence, our required answer $= 60-28=32.$

by Loyal (7.7k points)
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@shaik

therefore you have to use this formula n(A U B U C) = n(A)+n(B)+n(C) - n(A ∩ B) - n(B ∩ C) - n( A ∩ C) + n(A  ∩ B ∩ C )

formula used above is same only...its way of writing it differently..

the selection which we are doing i.e 3C1(n(A)+n(B)+n(C)) is too consider for all 3 elements

same for two elment case we  are doing 3C2..

and answer is 32 i guess its correct..!

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@ayush

yes in second case it should be 3P1 not 4C1...your answer is correct..!

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