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?