JEE Main 2015 · 11 Apr · Shift 1 · Mathematics Q61

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MathematicsSingle correct+4 / -1Question 61 of 90

Let $A=\{x_1,x_2,\dots,x_7\}$ 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 :

14\cdot{}^7C_2

16\cdot{}^7C_3

12\cdot{}^7C_2

14\cdot{}^7C_3

Answer and worked solution

Official answer: D

Choose the three elements mapped to $y_2$ : ${}^7C_3$ ways. The other four must map onto $\{y_1,y_3\}$ : $2^4-2=14$ ways. Total $14\cdot{}^7C_3$ .

Topics:Types of functions Combinations

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JEE Main 2015 · 11 Apr · Shift 1, Mathematics question 61