Questions $6,7$ and $8$ are interrelated and aim to see how Boolean algebra allows us to obtain simple combinational circuits.

Draw the circuit that implements the minimized Boolean expression obtained in question $6$ and indicate how many AND, OR and INV gates are required. There is no

restriction in the number of inputs of the AND and OR gates, use gates with as many inputs as you need.

Write the answer as $3$ integer numbers separated by a whitespace, in this order: number of AND gates, number of OR gates and number of inverters. For instance, if

$4$ AND gates, $1$ OR and $3$ inverters were required, the answer should be $413$

See how this circuit, equivalent to the one in question $6,$ requires fewer gates than the original.