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A $3$ line to $8$ line Decoder is used to implement a $3$-variable Boolean function as shown in figure.

The simplified form of output $Y$ is.

  1. $\bar{X}Y + \bar{Y}Z + XY\bar{Z}$
  2. $\bar{X}Z +\bar{Y}Z + XYZ$
  3. $X\bar{Y} + X\bar{Z} +\bar{X}YZ$
  4. $X\bar{Y} + X\bar{Z} + \bar{X}Y\bar{Z}$

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all bubble take ahead of OR gate that becomes NAND gate....

inputs to NAND gate are(1,3,5,6) which gives same output...using k-map...ans is C...
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HERE its written like this(Z is MSB) 

ZYX ....

NOW TAKE 0's outside and it will be NAND GATE now it says when enteries are either (1,3,5,6) we get 1

so fill 1 at those places and it will turn out to be Z'Y'X+Z'YX+ZY'X+ZYX' THAT MATCHES WITH c 

JUST MAKE THE OPTION C IN CANONICAL FORM

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The output would be 1, which is not given in the options.

If C would be the option then the decoder would form an AND-OR/NAND-NAND structure.

which means, For the C option, (1,3,5,6) shall be max term which they(ACE) have missed out.

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Please explain in detail