The number of surjective functions from \[A\] to \[B\] where \[A=\{1,2,3,4\}\] and \[B=\{a,b\}\] is
A. 14
B. 12
C. 5
D. 15
Answer
614.1k+ views
Hint: In the given question, we are given two sets namely, A and B and using these given sets we have to find the number of surjective functions. To calculate the number of surjective function, we will be using the formula, \[\sum\limits_{r=1}^{n}{{{(-1)}^{n-r}}^{n}{{C}_{r}}{{r}^{m}}}\]. Substituting the values of \[m=4\] and \[n=2\] in the given expression, we will get the value of the number of surjective functions.
Complete step by step solution:
According to the given question, we are given two sets namely, A and B and using these sets given to us we have to find the number of surjective functions.
Surjective function can be defined as a function f from a set X to a set Y, if every element in Y (codomain) has at least one element in the X. Uniqueness is not a necessity in this case.
The formula for finding the number of surjective function is, \[\sum\limits_{r=1}^{n}{{{(-1)}^{n-r}}^{n}{{C}_{r}}{{r}^{m}}}\]
Where m and n are the number of the elements of the sets X and Y respectively such that \[1\le n\le m\].
So, from the given values we can write,
The number of elements in set A is \[m=4\].
And the number of elements in set B is \[n=2\].
The condition is also fulfilled, that is, \[1\le 2\le 4\].
So, substituting the known values in the formula, we get the expression as,
\[\sum\limits_{r=1}^{2}{{{(-1)}^{2-r}}^{2}{{C}_{r}}{{r}^{4}}}\]
Expanding the above expression further and solving it, we get the value as,
\[= {{(-1)}^{2-1}}{{.}^{2}}{{C}_{1}}{{(1)}^{4}}+{{(-1)}^{2-2}}{{.}^{2}}{{C}_{2}}{{(2)}^{4}}\]
Using the formula of the combination to expand the expression, we have,
\[= {{(-1)}^{1}}.\dfrac{2!}{1!\left( 2-1 \right)!}(1)+{{(-1)}^{0}}.\dfrac{2!}{2!\left( 2-2 \right)!}(16)\]
Solving further, we get,
\[= (-1).\dfrac{2!}{1!1!}(1)+\dfrac{2!}{2!\left( 0 \right)!}(16)\]
\[= (-1).2!+1.(16)\]
\[= -2+16\]
So, we get the value as,
\[= 14\]
So, the correct answer is “Option A”.
Note: The formula to find the number of surjective functions between two sets has many components to it and so has a very good chance of writing the expression incorrectly. Also, the combinations formula applied in the above solution should also be carefully done, which is, \[\dfrac{n!}{r!\left( n-r \right)!}\]. In the above solution, we get \[0!\], it does not mean 0 rather its value is 1, so don’t get confused here and write the wrong value.
Complete step by step solution:
According to the given question, we are given two sets namely, A and B and using these sets given to us we have to find the number of surjective functions.
Surjective function can be defined as a function f from a set X to a set Y, if every element in Y (codomain) has at least one element in the X. Uniqueness is not a necessity in this case.
The formula for finding the number of surjective function is, \[\sum\limits_{r=1}^{n}{{{(-1)}^{n-r}}^{n}{{C}_{r}}{{r}^{m}}}\]
Where m and n are the number of the elements of the sets X and Y respectively such that \[1\le n\le m\].
So, from the given values we can write,
The number of elements in set A is \[m=4\].
And the number of elements in set B is \[n=2\].
The condition is also fulfilled, that is, \[1\le 2\le 4\].
So, substituting the known values in the formula, we get the expression as,
\[\sum\limits_{r=1}^{2}{{{(-1)}^{2-r}}^{2}{{C}_{r}}{{r}^{4}}}\]
Expanding the above expression further and solving it, we get the value as,
\[= {{(-1)}^{2-1}}{{.}^{2}}{{C}_{1}}{{(1)}^{4}}+{{(-1)}^{2-2}}{{.}^{2}}{{C}_{2}}{{(2)}^{4}}\]
Using the formula of the combination to expand the expression, we have,
\[= {{(-1)}^{1}}.\dfrac{2!}{1!\left( 2-1 \right)!}(1)+{{(-1)}^{0}}.\dfrac{2!}{2!\left( 2-2 \right)!}(16)\]
Solving further, we get,
\[= (-1).\dfrac{2!}{1!1!}(1)+\dfrac{2!}{2!\left( 0 \right)!}(16)\]
\[= (-1).2!+1.(16)\]
\[= -2+16\]
So, we get the value as,
\[= 14\]
So, the correct answer is “Option A”.
Note: The formula to find the number of surjective functions between two sets has many components to it and so has a very good chance of writing the expression incorrectly. Also, the combinations formula applied in the above solution should also be carefully done, which is, \[\dfrac{n!}{r!\left( n-r \right)!}\]. In the above solution, we get \[0!\], it does not mean 0 rather its value is 1, so don’t get confused here and write the wrong value.
Recently Updated Pages
Lysosomes are known as suicidal bags of cell why class 11 biology CBSE

Father s age is three times the sum of the ages of-class-11-maths-CBSE

Give a comparative account of the classes of kingdom class 11 biology CBSE

The ceiling of a long hall is 25m high What is the class 11 physics CBSE

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Name the Largest and the Smallest Cell in the Human Body ?

Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

10 examples of friction in our daily life

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Difference between physical and chemical change class 11 chemistry CBSE

