Simplifying a relay network
first:

the relays you've got can't wired directly to a arduino.

the data sheet isn't doesn't tell what current it draws / peek current not even the coil resistance or the build in resistor in series with the coil.

outgoing form the average power consumption of the coil divided by the max arduino output voltage delivers the lowest average current

P = V * I ==> I = P / U --> I = .29W / 5V = 58mA

the maximum current drawn or shucked to a IO is generally ± 25mA
the absolute maximum generally ±40 - 50mA

maybe the arduino survives the current of 58mA but it can die,
here you can find more info about not exceeding the max current.

also always put a diode across the relay coil to prevent frying the micro controller from the reverse voltage spikes coils generate when they are unloaded

it is not deadly if you don't put the diode, because the arduino outputs are push-pull; which means that that a output is shorted 5V (high) or shorted to gnd(0V) (low). So the coil is never be unloaded / floating. (but for transistors is it needed).



something like this, the transistor can handle more current than a relay.
the 12V in your case can variate between 5V and 24V.

the transistor type can be any type where 


max Vce higher is than the supply voltage (the 5 to 24V) (almost any common transistor is can handle that)
max IC is higher than 100mA
max Power doesn't care because the transistor will act like a switch


by example the BC547B is an excellent very common transistor
and has a gain (HFE) about 200 - 250.

the resistor value can be calculated with ohms law.

R = V / I


V = (5V -.07V) 5V is the max output voltage of the arduino, 0.7V is the voltage drop in the BE of the transistor.
I = Ic / HFe


IC = the max current drawn by the relay the previous maths says 58mA but I 'm not sure if more or less. thus i take 100mA 
HFe is the HFe of the transistor 200



I = 100mA / 200 = .5mA = 500μA

R = 5V / 500μA = 10kΩ

important

the HFe is very dependent from time, temperature, even the collector current and base current may affect the HFe.

so to be sure mostly the resistor value is divided by 2 or more.

but because we've taken the current almost twice as high, and the minimum HFe (the BC547b's HFe is between 200 and 450) and it is not a disaster is the transistor is not fully saturated but maybe 90% (VCE about 1V in stead of 0.3V) and 10kΩ a very common resistor we can take 10kΩ but if you want to be really safe, you can take a 5.6kΩ or less.

Second:

multiplexing


  there are several ways to multiplex
  very common is matrixing:
  
  
  
  just replace every led by a relay
  
  how it works
  
  first select a column than the row 
  
  the image is for demonstrating only, don't care about pin numbers
  
  connect each column pin and row trough a transistor / transistor chip
  to the arduino
  if you want 48 relays then a 6 by 8 matrix may do the trick
  
  there exist ICs that can do something similar 


More IO


  
  there exist io expanders over I2C or SPI
  with a lot of libraries for arduino
  
  generally a io expander acts like a io on the arduino and can handle about the same current maybe more or less
  bcd to decimal 
  converts a binary number to a single high output
  
  the 74HC42 (by example) can do this
  
  I don't know if there exist a library for a arduino. but if necessary, I can write you a code example that works (or a library is also possible)
  
  
  shift registers
  explained on the arduino site shiftout()
  
  the 74hc595 can do this