Thread: permutation question

If you have a five bit string ABCDE, how many strings can you make that have A before C and C before E?

Originally Posted by Frostking

If you have a five bit string ABCDE, how many strings can you make that have A before C and C before E?

In any rearrangement of “ABCDE” we can leave the B & D fixed rearrange the A, C & E is six ways. But only one of those do A, C & E appear in that order. So the answer is one-sixth of the total.

So, you are saying since there are 120 total ways to arrange a five bit string with ABCDE, there would be 20 of these in which A is before C and C is before E? Can I then think of it as choosing the other two members D and B in 5 x 4 ways? Or is that in error? Thanks so much for your prompt help!

since there got 5 string, let this to be 5 empty space
_ _ _ _ _
given situation some thing look like A->C->E (A before C before E)
from this 5 empty space, choose 3 of this 5, therefore there have C(5,3)=10
ways to put A C E, this left B and D, after finish put A C E, there left 2 free space for B and D. Hence C(5,3)*2*1=20

Hello, Frostking!

If you have a five bit string ABCDE, how many strings can you make
that have A before C and C before E?

Since A, C, E will appear in alphabetical order,
. . the only issue is the placement of B and D.

And there are:. ways.

Thank you Soroban and Lekge for adding your explanations. I really appreciate the help!!!!!