Test Series

Question

Let there are 12 minterms in a function in which 8 minterms are covered by 2 Essential Prime Implicants. Each of the remaining 4 mintermshave 2 Non- Essential Prime Implicants. Then the total number of minimal expressions is

a)4

b)8

c)16

d)1

 

can someone pls explain this with an example ? answer is 16.

Comments

Someone pls answer this ..

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8 min terms are covered by essential prime implicants means essential prime implecants must be choose we dont have any choice now other 4 minters we have 2 choice for each minterms so, ans is 2*2*2*2= 16

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Isn't 16 the number of ways we select the remaining 4 min terms ? What about the 8 mimterms that are covered by two essential prime implicants ?

Since we have to find out total number of minimal expression .. won't those two essential prime implicants be included .. ?

Kindly clear this doubt ..

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essential prime implicants should be included u dont have choice but non essential prime implicants have choice among 2 non prime implicants.

for 8 mintermscoverd by essential prime implicant we have nly 1 choice so

1*1*1*1*1*1*1*1*2*2*2*2

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Ok got it , thanks a lot :)