JEE Main 2015 (Online) 11th AprilQuestion 1
Let $A=\{x_1,x_2,\ldots,x_7\}$ 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 :
Permutation and Combination
Solution
Concept. Choose the $3$ elements going to $y_2$; the other $4$ must cover both $y_1$ and $y_3$. Formulas used. (1) Onto maps from a $4$-set to a $2$-set: $2^4-2$Step 1 - Preimage of $y_2$. ${}^7C_3$ ways. Step 2 - Other 4 elements. Each goes to $y_1$ or $y_3$: $2^4=16$; remove all-$y_1$ and all-$y_3$: $14$. Step 3 - Total. $14\cdot{}^7C_3$. Answer: C