Question

Let A=\left\\{ {{x}_{1}},{{x}_{2}},{{x}_{3}}.....{{x}_{7}} \right\\},B=\left\\{ {{y}_{1}},{{y}_{2}},{{y}_{3}} \right\\}. The total number of functions $f:A\to B$ that are onto and there are exactly three-element x in A such that $f\left( x \right)={{y}_{2}}$ is equal to
(a) 490
(b) 510
(c) 630
(d) None of these

Answer 1

Hint: In order to solve this question, we should have some knowledge of onto functions, that is, for a function $A\to B$, if all elements of B have at least one element matching with A, then the function is said to be onto function. Also, we need to know that whenever we have to choose r out of n elements, then we apply a formula of combination that is $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.

Complete step-by-step answer:
In this question, we have been asked to find the total number of functions $f:A\to B$, where A=\left\\{ {{x}_{1}},{{x}_{2}},{{x}_{3}}.....{{x}_{7}} \right\\},B=\left\\{ {{y}_{1}},{{y}_{2}},{{y}_{3}} \right\\}, that are onto and there are exactly three elements in A such that $f\left( x \right)={{y}_{2}}$. To solve this question, we should know that onto functions are those, for $f:A\to B$, all the elements B will have at least one element in A. And, we also know that for choosing r out of n items, we apply the formula of combination, that is,
$^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Now, we have been given that there are 3 elements in A which belongs to ${{y}_{2}}$ elements of set B. So, to choose 3 elements which belong to ${{y}_{2}}$ from A can be chosen by the formula of combination for n = 7 because A has 7 elements and r = 3 because 3 elements of A belongs to ${{y}_{2}}$, we get,
$^{7}{{C}_{3}}=\dfrac{7!}{3!\left( 7-3 \right)!}$
$^{7}{{C}_{3}}=\dfrac{7!}{3!4!}$
$^{7}{{C}_{3}}=\dfrac{7\times 6\times 5}{3\times 2\times 1}$
$^{7}{{C}_{3}}=35$
Now, after choosing 3 elements from A, we have been left with 4 elements that will either belong to ${{y}_{1}}\text{ or }{{y}_{3}}$. So, for each of the 4 elements, there are 2 possible ways to satisfy $f:A\to B$. So, the total number of ways of 4 elements of A having image with ${{y}_{1}}\text{ or }{{y}_{3}}$ is ${{2}^{4}}$. Now, we know that these cases will include those cases in all 4 elements that will have the image as either ${{y}_{1}}\text{ or }{{y}_{3}}$ which dissatisfies the condition of onto, that are 2 cases. And we want to generate an onto function, so to find the number of ways of creating onto function, we will subtract 2 from ${{2}^{4}}$. So, we get,
$={{2}^{4}}-2$
$=16-2$
$=14$
Hence, the total number of functions $f:A\to B$, that are onto and there are exactly 3 elements is A which belongs to $f\left( x \right)={{y}_{2}}$ are
$=14\times 35$
$=490$
Hence, option (a) is the right answer.

Note: While solving the question, the possible mistake one can make is a calculation or by not subtracting 2 from ${{2}^{4}}$, which are the number of ways of all elements belonging to any one of ${{y}_{1}}\text{ or }{{y}_{3}}$ and all possible ways of combining the remaining 4 elements to find ${{y}_{1}}\text{ or }{{y}_{3}}$ respectively will give us the wrong answer.