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.