Let A = \ x _ 1 , x _ 2 , , x _ 7 \ andB = \ y _ 1 , y _ 2 , y _ 3...

$A = { {x}_{1}, {x}_{2} \ldots . . {x}_{7} }$ and $B = { {y}_{1}, {y}_{2}, {y}_{3} }$

Let us select $3$ elements from $A$ & connect it to ${y}_{2}$

Number of ways $= {}^{7}{C}_{3}$

The number of ways would be equal to ${}^{7}{C}_{3}$ and now we have 2 elements ${y}_{1} \text{ and } {y}_{3}$ 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 ${y}_{3} )$ $= {2}^{4} - 2 = 14$

$∴$So the total number of onto functions would be $14{\cdot}^{7}{C}_{3}$