Permutations

[PaulStat]

So been a while since I've done anything like this. Say you have a 2 * 2 matrix where it's indices values can be 1 or 0. How many unique permutations are there?

Eg.

11
11

10
11

I started by drawing them out and got to about 13 or 14 then stopped to think if there was some formula which could work it out?

 

[bremen1874]

Not given this a great deal of thought, but isn't this just straight forward binary?

Flatten the grid into a binary value with the same number of columns as there are cell values and you should have the same thing.

1 = 1
11 = 3
1111 = 15
111111 = 63

and add 1 to account for the zero state

 

[adolf hamster]

isn't it a factorial?

if you take it as a row of xxxx then it's 1x2x3x4 which is 24 different combinations

 

[bremen1874]

isn't it a factorial?

if you take it as a row of xxxx then it's 1x2x3x4 which is 24 different combinations

xxxx
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111

only 16 options

 

[D.P.]

The matrix is meaningless, you have 4 binary values. 2^4=16

 

[adolf hamster]

xxxx
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111

only 16 options

not a factorial then

 

[thenewoc]

number of options multiplied by number of options and so on for as many positions you wish to calculate.

OP example is 2 x 2 = 4

00
11
10
01

 

[bremen1874]

number of options multiplied by number of options and so on for as many positions you wish to calculate.

OP example is 2 x 2 = 4

00
11
10
01

Try that with a 4 x 4 grid.

4 x 4 = 16

actual number of options = 65,536

 

[thenewoc]

Try that with a 4 x 4 grid.

4 x 4 = 16

actual number of options =

4 x 4 x 4 x 4 = 256

 

[bremen1874]

4 x 4 x 4 x 4 = 256

No even close

 

[thenewoc]

if there's 4 columns and each column can only be 0, 1, 2 or 3 then I'm sure 4 x 4 x 4 x 4 = 256 is correct.

 

[bremen1874]

256 would be 11111111 in binary (allowing for the zero case) which would be a 2 x 4 grid.

We may be at crossed purposes. If you read the open post only two values are allowed (0 and 1).

 

[thenewoc]

Actually I made a mishap, OP says you can only have 0 or 1 so it would be 2 x 2 x 2 x 2 = 16.

 

[PaulStat]

Reading this 16 looks like the right answer. Now I feel dumb

 

[LuckyBenski]

It's literally:

(Number of possible values the slot can have)^(number of slots)

0 and 1 gives you 2 options. 2^4 for a 2x2 grid allows 16 variants.

If it were, say, a password that could be lowercase a-z, uppercase A-Z plus digits 0-9,

26 + 26 + 10 = 62 possible values.

For a 4-character password, 62^4 = 14,776,336 possible passwords. Brute force cracker will get through that in no time.

Am I helping or making it worse