JEE Mains · Maths · STD 12 - 1. relation and function
Let \(A = \{ {x_1},\,{x_2},\,............,{x_7}\} \) and \(B = \{ {y_1},\,{y_2},\,{y_3}\} \) be two sets containing seven and three distinct elements respectively. Then the total number of functions \(f : A \to B\) that are onto, if there exist exactly three elements \(x\) in \(A\) such that \(f(x)\, = y_2\), is equal to
- A \(14.{}^7{C_3}\)
- B \(16.{}^7{C_3}\)
- C \(14.{}^7{C_2}\)
- D \(12.{}^7{C_2}\)
Answer & Solution
Correct Answer
(A) \(14.{}^7{C_3}\)
Step-by-step Solution
Detailed explanation
Number of onto function such that exactly three elements in \(x \in A\) such that \(f\left( x \right) = \frac{1}{2}\) is equal to \( = {\,^7}{C_3},\left\{ {{2^4} - 2} \right\} = 14.{\,^7}{C_3}\)
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