Am I doing something wrong? *SOLVED*

[her209]

Given four two-bit binary numbers

00
01
10
11

How many six-bit binary number permutations can be constructed?

That would be 4 P 3 = 4! / (4-3)! = 24.

But 2^6 = 64.

What did I go wrong?

 

[BlueWeasel]

The answer is 42.

 

[SWScorch]

2^6 implies you can arrange the bits any way you see fit. But with 4 2-bit numbers, those two bits always stay side-by-side, so it's not really 6 independent bits.

 

[SomeDude22]

remember .999999999999999999999999999999999999999999999999999999999999999999999999999999
999=1

 

[chuckywang]

for six bits, you need 3 of those two bit combinations. hence, 4^3 = 64. This is a stupid way of just doing the 2^6 that you had in the OP.

 

[walla]

you must assume you can't pick the same two bit number twice.

For instance, 111111 is not a valid choice.

Therefore, you won't have every 6-bit enumeration.

 

[ryzmah]

It depends if reuse is allowed - is 000000 a valid choice?

Without reuse, you have 4*3*2 = 24 arrangements.
With reuse, you have 4*4*4 = 64 arrangements.

 

[her209]

Originally posted by: SWScorch
2^6 implies you can arrange the bits any way you see fit. But with 4 2-bit numbers, those two bits always stay side-by-side, so it's not really 6 independent bits.

After thinking about your response some, I know where I went wrong. Using permutation, you can't get combinations that use repititive pairs, e.g.

000000
000001
000010
000011
... etc.

 

[JujuFish]

Originally posted by: SomeDude22
remember .999999999999999999999999999999999999999999999999999999999999999999999999999999
999=1

No. .999999999999999999999999999999999999999999999999999999999999999999999999999999
999=.999999999999999999999999999999999999999999999999999999999999999999999999999999
999 Duh!