The number of distinct combinations that can be created with n bits is 2n.
32 Bits 4 Octets with 1 Byte each(8 Bits)
5 bits
A 264-bit system has 64 bits.
4 bytes are equivalent to 32 bits.
How many bits are there in a data link layer ethernet address?
With two bits, there are (2^2) possible combinations, which equals 4. The combinations are: 00, 01, 10, and 11. Each bit can be either 0 or 1, leading to these four distinct configurations.
4 bits. 24 = 16, so you have 16 different combinations.4 bits. 24 = 16, so you have 16 different combinations.4 bits. 24 = 16, so you have 16 different combinations.4 bits. 24 = 16, so you have 16 different combinations.
26 = 64
Each bit allows 21 combinations. Accordingly, n number of bits provide 2n combinations.Therefore, 5 bits provide 25 = 32 combinations. # 00001 # 00010 # 00011 # 00100 # 00101 # 00110 # 00111 # ...
A 10-bit binary number can represent (2^{10}) different combinations. This is because each bit can be either 0 or 1, leading to (2) choices for each of the (10) bits. Therefore, (2^{10} = 1024) different combinations can be represented by 10 bits.
In a binary system, each bit can be either 0 or 1. Therefore, for 5 bits, the total number of combinations can be calculated as (2^5). This results in 32 different combinations, ranging from 00000 to 11111.
There are only 5 distinct combinations of 4 numbers.(1234, 1235, 1245, 1345, 2345) C = 5! / 4!But there are 120 distinct combinations in distinct order (i.e. 24 ways to order each abcd).abcdabdcacbdacdbadbcadcb
8192
For every bit, there are two combinations. For 12 bits, you need to multiply all together: 2 x 2 x 2 x ... x 2, or 2 to the power 12, which is equal to 4096.
32 values. 2^5=32
There are infinite combinations that can make 3879
9!/6!, if the six different orders of any 3 digits are considered distinct combinations.