Question

Let $X=\left\{ {{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}},{{a}_{5}},{{a}_{6}} \right\}$ & $Y=\left\{ {{b}_{1}},{{b}_{2}},{{b}_{3}} \right\}$ the number of function f from X to Y such that it is onto an there are exactly three elements x in X such that $f\left( x \right)={{b}_{1}}$ is
(a) 75
(b) 100
(c) 120
(d) 50

Answer 1

Hint: We start solving the problem by recalling the definition of onto function. We then find the number of ways to choose three elements from six elements of X by using the fact that the number of ways of choosing ‘r’ elements out of ‘n’ $\left( n\ge r \right)$ is ${}^{n}{{C}_{r}}$ ways. We then find the number of ways that the remaining elements of X map with the remaining elements of Y. We then multiply both the obtained ways to get the required number of functions satisfying the given condition.

Complete step by step answer:
According to the problem, we are given the sets $X=\left\{ {{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}},{{a}_{5}},{{a}_{6}} \right\}$ & $Y=\left\{ {{b}_{1}},{{b}_{2}},{{b}_{3}} \right\}$. We need to find the number function f from X to Y such that it is onto and there are exactly three elements x in X such that $f\left( x \right)={{b}_{1}}$.
We know that the function is said to be onto if every element in the co-domain is the image of at least one of the elements in its domain.
So, this means that ${{b}_{1}}$, ${{b}_{2}}$ and ${{b}_{3}}$ must have at least one pre-image in X.
But we need to find the onto functions such that there are exactly three elements x in X such that $f\left( x \right)={{b}_{1}}$.
So, let us first find the total number of ways to select three elements from six elements of X.
We know that the number of ways of choosing ‘r’ elements out of ‘n’ $\left( n\ge r \right)$ is ${}^{n}{{C}_{r}}$ ways. We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ......\times 3\times 2\times 1$.
So, we get the total number of ways of choosing three elements from X as ${}^{6}{{C}_{3}}=\dfrac{6!}{3!3!}=\dfrac{6\times 5\times 4}{3\times 2\times 1}=20$ ways ---(1).
Now, we need to map the remaining three elements from X to the remaining two elements in Y.
We know that each of the elements have 2 ways to map. So, the total number of ways to map is $2\times 2\times 2=8$. But we need to eliminate the mapping that all the elements map either to ${{b}_{2}}$ or ${{b}_{3}}$ which is two ways. So, the total number of ways will be $8-2=6$ ways ---(2).
Now, we need to multiply the results in equation (1) and (2) to get the total number of onto functions satisfying the given condition (multiplication is done as both processes have to be done simultaneously).
So, the number of functions will be $20\times 6=120$.

So, the correct answer is “Option c”.

Note: We should know that the element in set X should only map with one element in the set Y otherwise it would not be categorized as function. We should not solve the problem by ignoring the ways that the remaining three elements of X map with only one element of Y which is the common mistake done by students. Similarly, we can expect problems to find the total number of one-one and bijective functions formed by the given sets.