computer network

Question

In class C network 3 bits are 3 bits are borrowed for subnetting host id parts.Total possible number of subnet masks are ______ ?

Comments

They are eliminating 000 , 111 which is too wrong to do these days , i consider...

it should be explicitly mentioned that noncontinuous subnet mask can be considered VALID ...

then only i am agree with 56

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Yes they should mention about Cont / Non Cont Subnets.

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I also got C(8,3), but got confused coz it is given 42

Answer 1

According to your question,In class C network ,we have 24 bit Network Part and 8 bit host part.Now3 bits are borrowed from Host part(logically it will be borrowed from the host part only to create a subnet).then if we follow convention then there is only 1 subnet mask possible (convention is to take 3 bit borrowed from host is from msb of host) .

subnet mask $\Rightarrow$$\underbrace{11111111.11111111.11111111.111}\underbrace{00000}$(24 bit+3 bit=27 bit network part and 5 bit host part)

If you go by convention,i.e only possible subnet mask is $11111111.11111111.11111111.11100000$

here is the reference

http://www.subnet-calculator.com/subnet.php?net_class=C

but if you don't follow the convention i.e select 3 bit '1' from any one of 8 bit's host part we have $\binom{8}{3}*\binom{3}{3}$(selected 5 bit from 8 bit and then selecting 3 bit from remaining 3 bit)=56 possible subnet mask

Comments

What is subnet mask?

Network part contain only "$1$" and hosty part only "$0$"!

How can "$001$","$010$","$011$" ,"$100$" "$101$","$110$", be a part of subnet mask?

only possiblity is "$111$" making it "$255.255.255.224$"

$\Rightarrow 11111111.11111111.11111111 .11100000$

Only by convention

Else

select "$111$" from 8 bits of host $\Rightarrow \binom{8}{3}=56$

Check it http://www.subnet-calculator.com/subnet.php?net_class=C

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@Sourav. Why cant "001","010","011" ,"100" "101","110" b part of the subnet mask?

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If you don't go by convention then 3 bit must be "$1$","$1$","$1$" because  3bit is made the network part which must be "$1$".

so mask can't be "$11111111.11111111.11111111.00100000$"

number of "$1$"s should be 24+3=27 but your's is 25

rather it can be 

"$11111111.11111111.11111111.00111000$"

(not by convention)

Answer 2

According to standard convention, there should be exactly 1 subnet mask.

It is a Class C network, therefore, 24 Most Significant bits are for Network ID part and Remaining 8 bits are for Host ID part. Now, in the question, 3 bits from Host ID part are taken for subnetting.

So Subnet Mask has to be 11111111.11111111.11111111.11100000

But, it is not mentioned as to which 3 bits of the 8 bits (Host ID part) are used for subnetting. So Mathematically, we have $^{8}C_{3}$ choices to make and thus $^{8}C_{3}$ = 56 subnet mask are possible.

Edit: The question does not mention about continuous/discontinues bit.  If use of continious bits are not allowed then answer will be less then 56.

The question should have been more described. I would have gone for 1 as answer had this been an actual GATE question.

Comments

So the answer given here http://www.techtud.com/example/class-c-address-3-bits-are-borrowed is wrong.

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The question does not mention about continuous/discontinues bit.  If there is restrictions imposed on use of continuous 1 then answer will change.

so, that answer is not wrong. 56 is also not wrong. 1 is also not wrong. It depends on how well described the question is.