edited by
722 views
0 votes
0 votes

The number of ways possible to form injective function from set A to set B where |A| = 3 and |B| = 5, such that pth element of set A cannot match with pth element of set B are _________.

can someone explain this question to me

edited by

1 Answer

Best answer
4 votes
4 votes
You can apply inclusion exclusion principle here.
Suppose P is define as Pth element of A matches with Pth element of B.
Then we have to find N(P1` P2` P3`) i.e none of the Pth element of A matches with Pth  element of B

According to inclusion exclusion principle

N(P1`P2`P3`) = N - [N(P1) + N(P2) + N(P3)] + [N(P1P2) + N(P1P3) + N(P2P3)] - [N(P1P2P3)]

N = Total number of injective funtion possible..
For above case |A| = 3 and |B| = 5, So N = 5 x  4 x 3 = 60

N(P1) = Total number of injective function such that first element of A matches with first element of B.
So N(P1) = 4 x 3 = 12
For N(P2), N(P3), we are going to get same result.

N(P1P2) = Total number of injective function such that first two element of A matches with first two element of B
So N(P1P2)  =  3
For N(P1P3) & N(P2P3), we are going to get same result.

N(P1P2P3) = Total number of injective function such that all three element of A matches with first three element of B
So N(P1P2P3) = 1

Therefore N(P1`P2`P3`) = 60 - [12 + 12 + 12] + [3 + 3 + 3] - [1] = 60 - 36 + 9 -1 = 32

So 32 is correct answer.
selected by

Related questions

1 votes
1 votes
0 answers
1
Chhotu asked Nov 26, 2017
498 views
Hi Guys,I think answer should be (D) part because it is not mentioned that function is s Injective. What is your opinion ?
1 votes
1 votes
0 answers
3
1 votes
1 votes
1 answer
4