JEE Main 2015 — Permutation Combination Question with Solution

$A=\{x_1, x_2\dots ..x_7\}$ and $B= \{y_1, y_2, y_3\}$

Let us select $3$ elements from $A$ & connect it to $y_{\mathrm{2 }}$

Number of ways $={}^7{C}_3$

The number of ways would be equal to ${}^7{C}_3$ and now we have 2 elements $y_1 \& y_3$ in set $B$ to be mapped from the remaining element of set $A$

Hence, total number of function $2^4=16$ and out of which $2$ would be "into" functions (when all four goes in $y_1$ & when all four goes in $y_3)$ $=2^4-2=14$

$\therefore$So the total number of onto functions would be $14{·}^7{C}_3$