architecture - Endian-ness: Bits in a Byte vs. Bytes in Memory -

when specific architecture either little-endian or big-endian, referring whether numerical significance stored left-to-right or right-to-left in memory. question is: ordering refer how bits or ordered in byte, or how bytes ordered in memory?

for example, consider number 6000=1770h=0001011101110000b. if both bits in byte , byte in memory little-endian, stored as

00001110 11101000 = 0e e8,

if bits in byte big-endian, bytes in memory little-endian, stored (for it's worth, happens how visual studio seems telling me memory organized in x64 architecture)

01110000 00010111 = 70 17, 

if bits little-endian, bytes big-endian, stored as

11101000 00001110 = 0e e8, 

and finally, if bits big-endian, bytes little-endian, stored as

00010111 01110000 = 17 70 

(hopefully did right.)

so then, terms "little-endian" , "big-endian" refer to? terms refer ordering of bits in byte, or ordering of bytes in memory, or both? furthermore, if vs tells me that, example, 7c, 'in' given particular byte, mean bits make byte in computer memory literally 0111 1100, or mean value stored in byte 7ch=124, or may not represented 7c=01111100 depending on whether or not underlying architecture happens little-endian?

the ordering of bits in byte invisible. since can't address individual bits, there no difference between 2 cases. however, can address individual bytes, there make difference.

if we're expressing 6000 in byte-addressible memory, high byte 23 decimal (6000 divided 256) , low byte 112 decimal (6000 mod 256). store 23,112 or 112,23. there no other options. ordering of bytes open choice, , endianness refers to.