There is 16 subnet being made when we extend /23 to /27.
Because we are using 4 bits from host bits.

2*2*2*2 = 16


so the subnets are :-

172.17.32.0 to 172.17.47.0

Thanks

There is 16 subnet being made when we extend /23 to /27.
Because we are using 4 bits from host bits.

2*2*2*2 = 16


so the subnets are :-

172.17.32.0 to 172.17.47.0


and the total number of host is 16x255 =4080 host

There is 16 subnet being made when we extend /23 to /27.
Because we are using 4 bits from host bits.

2*2*2*2 = 16


so the subnets are :-

172.17.32.0 to 172.17.32.31

172.17.32.32 to 172.17.32.63

and continue...with 32 IPs per subnet...

172.17.32.0/23
mask of that ip is              
sunated in the 
172.17.32.0/27
sunating of 4 bit
total number of n/w
2^4-2=14(according to cisco)
now taol nomber of valabe bit for host is 16-11=5
so taotal nomber of avilable host is 2^5-2=30
note**(why eleven is redused from 16.ans becoz this is a 
xlass b adress and it have 16 bit for n/w and 16 bit for 
host but we have the mask of 27 then we have taken 11 bit 
from host for n/w.so we redused that eleven bit from the 16)
now
now hat is the mask of that ip 
255.255.128+64+32+16+8+4+2+1.128+64+32
i.e
mask is 255.255.255.224/27
so the block size is 256-224=32
so the n/w is 
172.17.32.0---------------------.31
.32------------------------------.63 and so on

172.17.32.0/23 into a /27 how mant subnet?

n=23-27=4

total no subnet mask =2^n - 2=2^4-2=14
where n is the no host 
host =2^n-2=14
subnet =2^n=2^4=16

Dealing with a 172 IP places you into the Class B Network. 
The default subnet mask is 255.255.0.0.  However, with the
classless identifiers, or 'prefix length', of /23 and /27
your subnet masks will be as follows:

/23  255.255.254.0
/27  255.255.255.224

Which is broken down into binary as follows:
Default 11111111.11111111.00000000.00000000
/23     11111111.11111111.11111110.00000000
/27     11111111.11111111.11111111.11100000

If you'll notice there are 23 ones in the second subnet mask
and 27 in the third.  Hence the /23 and /27.  Pay attention
to the 'overlap' (area where the default reads 0 and the
prefix length reads a 1. In this case the last two octets) 
For the /23 there are 7 and the /27 there are 11.  Remember
these as they will come up later.

Note: There is a  proper term for the 'overlap' but I don't
remember it

Beginning at the right of the binary version of the /23
subnet mask. Start with the number 1 and double it for every
'0' you have, stopping ON the first '1'.  i.g.

                             512 256 128 64 32 16 8 4 2 1  
/23  11111111.11111111.111111 1   0.  0   0  0  0 0 0 0 0
/27  11111111.11111111.111111 1   1.  1   1  1  0 0 0 0 0

As you can see for the /23 subnet mask, 512 is the TOTAL
number of hosts you can have per subnet.  You'll have to
subtract 2 hosts for the network and broadcast addresses
leaving you with 510 USABLE hosts per subnet.  So, both look
like this:

Prefix           TOTAL            USABLE
Length           Subnets          Subnets
/23              512              510
/27              32               30

Remember the 'overlap' areas from above?  if not I'll give
them to you again:

/23  7
/27  11

This number will be used to find out the number of subnets
by using the following formula:

2^n
Where n is the the 'overlap'.

So...
 
2^7= 2x2x2x2x2x2x2=128 or
2x2=4x2=8x2=16x2=32x2=64x2=128
1 2   3   4    5    6    7

2^11= 2x2x2x2x2x2x2x2x2x2x2=2048 or
2x2=4x2=8x2=16x2=32x2=64x2=128x2=256x2=512x2=1028x2=2048
1 2   3   4    5    6    7     8     9    10     11

Like with the TOTAL hosts, we have to remove 2 from the
TOTAL subnets for network and broadcast ranges.

Now we know:

Prefix        TOTAL         USABLE
Length        Subnets       Subnets
/23            128           126
/27           2048          2046

Lets compile everything:

Prefix        USABLE        USABLE         Total Network
Length        Subnets       Hosts              HOSTS
/23            126          510               64,260
/27           2046           30               61,380

So, even though we gain 1920 subnets.  We lose 480 hosts per
subnet.  Which means we will lose a total potential of 2880
hosts.

I have one bit of clarification on my answer.  As the IP
address of, 172.17.32.0, was given that will reduce the
amount of subnets.  The third octet (.32.) has restricted us
to one subnet on the /23 prefix.  so that will change the
total subnets for the /27, see the following:

Prefix               Subnets in              Hosts per
Length               Range Given             Subnet
/23                  1                       510
/27                  14                      30

So, the Subnets will increase by 13 and the hosts will drop
to 420 (14x30).  Hope I got it this time.

Hi all i read all answers and in every answer you have some mistake.IT ask how many subnets and how many hosts have per subnet not how many host have totally!
So here is the most logical and easy understanding of sub-netting  with out need to convert to binary we all know to make binary no. from the regular number.
23 subnet mask =255.255.254.0
27 subnet mask =255.255.255.224
it ask for 27 how many subnets will have and how many  hosts
 so we start
we all should know that totally in subnet mask have 32-bits here first  27 are taken for the subnet mask so we left 5 bits for host 32-27=5
number of the host we get from formula 
2^n-2=No of host per subnet
so we get 
2^5-2=32-2=30
we take 2 address for the broadcast and subnet thats why is 2^n-2
We finish with  the hosts 
no lets get the no. of subnets
first given was /23 bits and the second was /27-bits
we just do this
27-23=4
no. of subnet is 2^n we don`t have here -2!!! some one make mistake in the answers so remember you don`t take minus 2 when you solve no. of subnets thats for hosts!!
so
27-23=4
2^4=16
ANSWER OF THE QUESTION IS:
Subnets 16
HOST 30 per subnet
address range like this
172.17.32.0 first subnet
172.17.32.32
172.17.32.64
172.17.32.96
172.17.32.128
172.17.32.160
172.17.32.192
172.17.32.224
172.17.33.0
172.17.33.32
172.17.33.64
172.17.33.96
172.17.33.128
172.17.33.160
172.17.33.192
172.17.33.224 last one
I hope i help to some of you:)
THIS IS TE REAL ANSWER!