Homework problem : Option B. Explaination : Actually the direct answer should be F = A'BC + B'AC + C'AB The avobe result can be directly interpreted from the ven diagram. But option is not available. So we have to solve the options. And bileve me other then options B no other options were related to the given Venn diagram, sir had just given them anonymously. Still I am going to prove how option B is correct. First of you all you can see E' had been ANDed (multiplied) to the correct answer . So you must have the concept of E E= A.B.C E'=A'+B'+C' Solving option B we have to prove A'BC + B'AC + C'AB F= (ABC' + A'BC + AB'C )E'. = (ABC' + A'BC + AB'C )(A'+B'+C') { just multiply(AND) each term} = ABC' A' + ABC' B' + ABC' C' + A'BCA' + A'BCB' + A'BCC' + AB'CA' + AB'CB' + AB'CC' { X.X'=0} =ABC' C' + A'BCA' + AB'CB {X'.X'=X'} =ABC' + A'BC + AB'C Which is the result. Easy Right . Hope you like my explaination.Please do comment below if you still have doubt
E' is universal set so value = 1 Directly from Venn diagram, we can write shaded region = (AB'C + A'BC + ABC') (AB'C + A'BC + ABC')•1 And E' = 1 So (AB'C + A'BC + ABC')E' Option b is correct
Answer is B How? Explained below... As we take the term where we see A and B meets and C is not included...we take it simply as ABC' but that's wrong....as E' is the entire region and not any particular shaded region so we would write it as ABC'E'(first term) Same thing would happen with all three terms... This will give Y=( ABC'E' + AB'CE' + A'BCE') take E' common out of all...and it will give us option B.
@@harshitagaur260 I am very sorry mam I just wanted to thank u for the explanation .sorry for the bad choice of words but it was totally unintentional . Sorry again
Firstly answer is (ABC'+AB'C+A'BC) But there is no option match E= A'B'C' E' = A+B+C BY making Boolean expression (ABC'+AB'C+A'BC).E' Gives the same result... just to match the option
option is b but what is the need of E in the exression as you are not shaded it... why should we use that in that in the expression?? can u pls explain..........
Guys don't make fuss just multiply your answer by E' because the whole expression is in E' itself. I think this makes your doubts clear don't go in the deep just think answer logically becaz this is logic electronics.
Option b is correct E'is universal set (A'.B.C+A.B'.C+A.B.C') REPRESENT SHADED REGION So universal set and shaded region gives required set of shaded region
CASE1_we have to Solve this for shaded region so Answer will be.... Y=AB'C+ABC'+A'BC CASE2_ But there Is E' which is equal to (A'B'C')' = A+B+C.... So Here I stuck... Sir plz Let me know Answer....🔜
I'm getting two expressions (A.B).C_bar+(B.C).A_bar+(C.A).B_bar and (A.B).(ABC_bar)+(B.C).(ABC_bar)+(C.A).(ABC_bar) if ABC=E then ABC_bar=E_bar. So, somebody please help me explaining whether my derivations are correct or not. If it's not then please help me with the correct one :) . Thanks in advance.
a is the correct answer because: the shaded part can be written as: (A.B).(A.B.C)'+(C.B).(A.B.C)'+(A.C).(A.B.C)' solving that we get : (A.C.B)'(A.B+B.C+C.A) and as we can see in the figure in the video (A.C.B)' is given as E' thus the ans is option a i.e E'(A.B+B.C+C.A)
JUST THINK LOGICALLY IN THIS WAY: F = [(B+C)-A] + [(A+C)-B] + [(A+B)-C] Meaning SHADED REGION =[ "whatever COMMON is their in B and C, rejecting part of A" + "whatever COMMON is their in A and C, rejecting part of B" + "whatever COMMON is their in A and B, rejecting part of C"] AND E' , Since E' is the universal set F =[ BCA'+ACB'+ABC'].E' [Here We can represent, (B+C)-A => BCA' ]
Homework question: Shade area: AB(ABC)' + BC(ABC)' + AC(ABC)' Further solving: (ABC)'[ AB + BC + AC ] #optionA Further solving by opening and multiply (ABC)' (A' + B' + C')(AB + BC + CA) //{A'.A = 0} A'BC + AB'C + ABC'
Does anyone understand why for (b) AB' + A + B is not correct? It seems to me that it should be correct, since if something is both A and B, then it also falls under the A + B. Am I wrong?
by the approach you solved, you will get : no result need not be unique, but since (A' + B) != (A + B') and for this particular question, there is only 1 possible minimization. So, your answer may be wrong. If you may provide me your approach or soln.
region is A'BC+AB'C+ABC' so as there s no answer matchin we have to solve the options....... and E=A'B'C' and E'=A+B+C and (A'BC+AB'C+ABC')(E')=A'BC+AB'C+ABC' SO PROVED AND using boolean formulaes
Could anyone explain to me why my boolean minimisation gives the wrong result? A’B’+AB’+AB = A’B’+A(B’+B) = A’B’ + A = A’+B’+A = A’ + A + B’ = 1 + B’ = 1
Neha wasnik you have to include all three variables in each term of the answer. The other 2 examples have 2 variables (2 circles), a & b. When just circle a was shaded in the first example he still included b, but complemented it. Therefore it was AB'. Now look at the third example. The top shaded region includes a & b, but not c. So it would be ABC'. If the middle region was shaded it would be ABC. Do you see? It's hard to explain in text lol
Homework problem : Option B.
Explaination :
Actually the direct answer should be
F = A'BC + B'AC + C'AB
The avobe result can be directly interpreted from the ven diagram. But option is not available.
So we have to solve the options.
And bileve me other then options B no other options were related to the given Venn diagram, sir had just given them anonymously.
Still I am going to prove how option B is correct.
First of you all you can see E' had been ANDed (multiplied) to the correct answer . So you must have the concept of E
E= A.B.C
E'=A'+B'+C'
Solving option B we have to prove
A'BC + B'AC + C'AB
F= (ABC' + A'BC + AB'C )E'.
= (ABC' + A'BC + AB'C )(A'+B'+C')
{ just multiply(AND) each term}
= ABC' A' + ABC' B' + ABC' C' + A'BCA' +
A'BCB' + A'BCC' + AB'CA' + AB'CB' + AB'CC'
{ X.X'=0}
=ABC' C' + A'BCA' + AB'CB {X'.X'=X'}
=ABC' + A'BC + AB'C
Which is the result. Easy Right .
Hope you like my explaination.Please do comment below if you still have doubt
prarthana roy Yes, you are correct . I didn't thought about it.
Readers please make a change in
E = A+B+C.
Before ans line the last term should be AB'CB'
Thanks bhai 😊
@@indranilsarmah5979 I think E'=A+B+C bro
@@indranilsarmah5979 here e means union of A B C WHICH means and of ABC SO IT IS CORRECT WAT U WROTE
Option B Because every region does not contain E so we take E' as common and the rest is clear from the respective shaded regions.
E' is universal set so value = 1
Directly from Venn diagram, we can write shaded region =
(AB'C + A'BC + ABC')
(AB'C + A'BC + ABC')•1
And E' = 1
So
(AB'C + A'BC + ABC')E'
Option b is correct
Answer is B
How?
Explained below...
As we take the term where we see A and B meets and C is not included...we take it simply as ABC' but that's wrong....as E' is the entire region and not any particular shaded region so we would write it as ABC'E'(first term)
Same thing would happen with all three terms...
This will give Y=( ABC'E' + AB'CE' + A'BCE')
take E' common out of all...and it will give us option B.
@@hashirwaqar8228 WTF
Thanks for the explanation.✌
@@harshitagaur260 just ignore him....he is just being an asshole...
@@harshitagaur260 I am very sorry mam I just wanted to thank u for the explanation .sorry for the bad choice of words but it was totally unintentional . Sorry again
@@oo7338 i really am sorry🙂
SIR I HAVE COMPLETED LECTURE 24 OUT OF 202. TREMENDOUS LECTURE AND THANKYOU FOR YOUR GUIDANCE !!
E’ is universal set. So that will come with each and every shaded region.
option B is the answer.
Firstly answer is (ABC'+AB'C+A'BC)
But there is no option match
E= A'B'C'
E' = A+B+C
BY making Boolean expression
(ABC'+AB'C+A'BC).E'
Gives the same result...
just to match the option
sir plz upload lectures on microprocessor and microcontroller,.,.
What is the significance of E' in the last question? Does it correspond with the universal set?
Yes
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Teaching skills is the best ❤❤❤❤
Sir, we could have simply used the De Morgan's law to minimize the shaded region expression.
option is b but what is the need of E in the exression as you are not shaded it...
why should we use that in that in the expression?? can u pls explain..........
Sir, kindly give the explanation for third question
I think option B is correct ....correct me if I'm wrong
Guys don't make fuss just multiply your answer by E' because the whole expression is in E' itself.
I think this makes your doubts clear don't go in the deep just think answer logically becaz this is
logic electronics.
Option b is correct
E'is universal set
(A'.B.C+A.B'.C+A.B.C') REPRESENT SHADED REGION
So universal set and shaded region gives required set of shaded region
Answer is option B and it's minimal form is (AB+C)E' but I really can't understand why the E' is multiplied.
P.S: I know I'm 6 years late 😂
I am still wondering how he I am not explaining in depth because u have being taught in school
8 years late 👋
@@MrGrey64 hehe
CASE1_we have to Solve this for shaded region so Answer will be.... Y=AB'C+ABC'+A'BC
CASE2_ But there Is E' which is equal to (A'B'C')' = A+B+C.... So Here I stuck... Sir plz Let me know Answer....🔜
Here E' = A' + B' + C' which means the whole space excluding A,B,C
Option A=(AB+BC+CA)Ē
Take E=A'.B'.C' FROM DIAGRAM
SO E'=A+B+C
Ans b is correct
Answer: B
I'm getting two expressions (A.B).C_bar+(B.C).A_bar+(C.A).B_bar
and (A.B).(ABC_bar)+(B.C).(ABC_bar)+(C.A).(ABC_bar) if ABC=E then ABC_bar=E_bar.
So, somebody please help me explaining whether my derivations are correct or not. If it's not then please help me with the correct one :) . Thanks in advance.
E_bar is the total region which is equal to 1?
Sir, can u explain d last one through ur comment?
I am confused😰.
E' stands for which region?
As per my analysis, E' is unshaded entire region
can you tell us the software you are using for teaching
don't need to solve entire thing as (A.1)=A so (A'BC + B'AC + C'AB ) . E' gives (A'BC + B'AC + C'AB ) itself here E' is universal set or 1
E' is not the universal set
for three variable venn diagram option B is right choice i think....am i right explain me
+Santhan Kumar yes u r right
Thank you sir❤❤❤❤
a is the correct answer because:
the shaded part can be written as: (A.B).(A.B.C)'+(C.B).(A.B.C)'+(A.C).(A.B.C)'
solving that we get : (A.C.B)'(A.B+B.C+C.A) and as we can see in the figure in the video (A.C.B)' is given as E' thus the ans is option a i.e E'(A.B+B.C+C.A)
JUST THINK LOGICALLY IN THIS WAY:
F = [(B+C)-A] + [(A+C)-B] + [(A+B)-C]
Meaning
SHADED REGION =[ "whatever COMMON is their in B and C, rejecting part of A" + "whatever COMMON is their in A and C, rejecting part of B" + "whatever COMMON is their in A and B, rejecting part of C"] AND E' , Since E' is the universal set
F =[ BCA'+ACB'+ABC'].E' [Here We can represent, (B+C)-A => BCA' ]
Arup Dey thanks a lot 😁😁😁
E' is not universal set
sir i m confused plz explain which option is correct
Bro that's fine, you r solving rasy question and you are giving hard questions to us thats not fare
Is he the first teacher you've seen?
B) Option is correct answer, but how E' is included
WE HAVE TO CONSIDER ALL THE VARIABLE PRESENT IN THE VENN DIAGRAM
ab intersect with E
Consider all the variables
Including e
Who is watching in 2024 😂
I love you so much ...and you are the best at all
Option B
a and b gives same answer but a looks much more simplified so I would go for a.
OPTION - A is correct (AB+BC+CA)
b option as E=A'B'C' so E'=A+B+C AND (ABC' +A'BC+AB'C)(A+B+C)= answer which is shaded region
Option A
Unlike in other examples why do we include E' here even though it is not a shaded portion! ! can anyone help me out! !
it is because E` is with all 3 A B and C
sir kindly give some explanation on propagation delay...sir plzz help me...
In the hw question E'=1.because it is universal .so ans will b.
Homework question:
Shade area:
AB(ABC)' + BC(ABC)' + AC(ABC)'
Further solving:
(ABC)'[ AB + BC + AC ] #optionA
Further solving by opening and multiply (ABC)'
(A' + B' + C')(AB + BC + CA)
//{A'.A = 0}
A'BC + AB'C + ABC'
E'=1 so b is correct because we anded with our expression
Does anyone understand why for (b) AB' + A + B is not correct? It seems to me that it should be correct, since if something is both A and B, then it also falls under the A + B. Am I wrong?
For the first example, when I simplify the expression I get A(complement )+B should the final result be unique ?
by the approach you solved, you will get : no result need not be unique, but since (A' + B) != (A + B') and for this particular question, there is only 1 possible minimization. So, your answer may be wrong.
If you may provide me your approach or soln.
why we are not using + operator so that A region can be written as a{comp}+b
They have asked sop
Answer is Option B
How?
Can u explain this?
region is A'BC+AB'C+ABC' so as there s no answer matchin we have to solve the options....... and E=A'B'C' and E'=A+B+C and (A'BC+AB'C+ABC')(E')=A'BC+AB'C+ABC' SO PROVED AND using boolean formulaes
Gurvir Singh how can u sat E' = a+b+c
SIE I DON'T KNOW HOW TO SOLVE THIS HOMEWORK QUESTION..........
sir i think question c for Venn diagram option b is correct if not means please explain it....
B option will be the correct one...
b is correct option
Can u please give explanation of hme wrk problem
Thanku sir B part would be Ans
option b in the brackets has the stright ans. again what is the reason of placing the ebar @Nesoacademy
Did u got your answer????
what is the region of E_bar
🔴H/W
✔️Ans: Option (B)
How plz let me know ,you are very active in maths aptitude also
Could anyone explain to me why my boolean minimisation gives the wrong result? A’B’+AB’+AB = A’B’+A(B’+B) = A’B’ + A = A’+B’+A = A’ + A + B’ = 1 + B’ = 1
because: A'B'+A = (A'+A) . (B'+A) = 1.(B'+A) = A+B' ------->maybe it is not useful after 1 year 😁
B is correct
I think the answer is option B. But here whats the importance of E'???
option (a) is the right answer
if you solve (ab+bc+ca).(a'+b'+c')
where a'+b'+c'=E'
you get the right one abc'+ab'c+a'bc
Sir plz give the derivation for minimization
my ans comes to match option 1
What is the answer of the question and how
Ans: b) ABC' + AB'C +A'BC
E' is not in shaded part ?
consider two two circles ....option A
Option B is correct
The answer is option B.
the answer is b
option b is the ANSWER
Answer is option B
what is the final answer?
sir please give me the explaination 3rd example
how E' is included?
E' = 1. The full area.
So including this will give the same answer
Answer is option(a)
A'BC+AB'C+ABC'
Ans is B
please explain h.w.problem anybody.
Ans is (b)
Ans B
Ans : Option B
Option (b)
option b will be thye correct answer
Sir b=B
I think ans is b. if wrong please explain
Nice. But voice is low
A is the answer
Wow dis is hard
give the answer with explain
OPTION A
option B
the answer is A for sure... just think people
its b lol
Its A middle region is not shaded
Neha wasnik you have to include all three variables in each term of the answer. The other 2 examples have 2 variables (2 circles), a & b. When just circle a was shaded in the first example he still included b, but complemented it. Therefore it was AB'. Now look at the third example. The top shaded region includes a & b, but not c. So it would be ABC'. If the middle region was shaded it would be ABC. Do you see? It's hard to explain in text lol
Neha wasnik loli b hoga😂
@@adamamaral5949 thanks a lot...u explained clearly
Option b
why E' ??
E'=1 as E'=a+b+c and a+b+c=1;
E+E' = 1 you are wrong
if E + E' = 1 , then option B wont be corrct.
here E' = should be 1
option b
It's option A
A