Writing A'B'CD+ABC' using two inverters and 5 2:1 multiplexers

Question

The question says draw F(A,B,C,D)=∑(3,7,11,12,13). I derived A'B'CD+ABC'. I am trying to draw it using two inverters and 5 2:1 multiplexers but i couldn't connect the output to the separate components i wrote. I know the correct answer but i just couldn't understand it.

Here's the correct solution

Why is the last mux connected to the 0 instead of 1 like we did all the other components? And why did they give 1 to 1 in mux in the answer?

Answer 1

OK, then maybe this will help:

F(A,B,C,D)=∑(3,7,11,12,13).

w/ 2 nots; 5 2/1 muxs

truth table
ABCD R
0000 0
0001 0
0010 0
0011 1
0100 0
0101 0
0110 0
0111 1
1000 0
1001 0
1010 0
1011 1
1100 1
1101 1
1110 0
1111 0

kmap
  \ CD 00 01 11 10
AB \ 
00     0  0  1  0
01     0  0  1  0
11     1  1  0  0
10     0  0  1  0

expression
ABC'+A'CD+B'CD

simplifying
ABC'+(A'+B')CD
ABC'+(A'+B')''CD
ABC'+(AB)'CD
(AB)'CD + ABC'

aux truth table:

(AB)'CD  ABC'  ((AB)'CD + ABC')
 0       0              0             see note 2
 0       1              1             see note 1
 1       0              1             see note 2
 1       1              1             see note 1

note 1: If ABC' is true (mux select is 1) then output is true (mux's 1 input is set to 1) note 2: If ABC' is false (mux select is 0) then output is (AB)'CD (mux's 0 input is set to (AB)'CD), the "see note 2" outputs are true only when (AB)'CD is true