Question
Question: Number of onto(surjective) function from A to B if \( n\left( A \right) = 6 \) and \( n\left( B \rig...
Number of onto(surjective) function from A to B if n(A)=6 and n(B)=3 is
A. 26−2
B. 36−3
C. 340
D. 540
Solution
In order to find the number of onto(surjective) functions between A and B , let us assume the n(A)=m and n(B)=n . Since 1⩽n⩽m , use the direct formula for number of surjective functions as ⇒∑r=1n(−1)n−rnCrrm by putting the values of m and n. Simplify the expression to get the required result.
Complete step by step answer:
We are given a relation A to B in which n(A)=6 and n(B)=3
n(A) basically denotes the number of elements in the A set and similarly n(B) denotes the number of elements in the B set. Let them be m and n respectively.
According to the question, we are supposed to calculate the number of onto or surjective functions that can be created between the elements of set A and B.
So surjective or Onto functions are the functions between A and B in which for every element of B, there is at least one or more than one element matching with A.
To calculate the number of such functions, there is a direct formula for 1⩽n⩽m
No of onto functions=
⇒∑r=1n(−1)n−rnCrrm
Putting m=6,n=3 , we have
⇒∑r=13(−1)3−r3Crr6 ⇒(−1)23C1(1)6+(−1)13C2(2)6+(−1)03C3(3)6
Remember the formula of C(n,r)=nCr=r!(n−r)!n!
Simplifying the expression further, we get
⇒(3)6−3×26+3=3((3)5−26+1) ⇒540
So, the correct answer is Option D.
Note: 1. Relation: Let A and B are two sets. Then a relation R from set A to set B is a subset of A×B
Thus, R is a relation from A to B ⇔R⊆A×B
If R is a relation form a non-void set A to a non-void set B and if (a,b)∈R ,then we write
aRb which is read as “a is related to b by the relation R “. If (a,b)∈/R , then we write aRb and we say that a is not related to b by the relation R.
2. The formula of number of surjective functions is only value for 1⩽n⩽m , if m<n , the number of onto functions is 0 as it is not possible to use all elements of B .