# Thread: Permutation of the letters ABCDEFGH

1. ## Permutation of the letters ABCDEFGH

Q.1 How many permutations of the letters ABCDEFGH contain

a) the strings CAB and BED?

myans.) 5P5 = 5! = 120 ways

b) the strings BCA and ABF?

myans.) permutation not possible since B cannot be followed by both C and F at the same time.

Thanks for any help!

2. Originally Posted by robocop_911

Q.1 How many permutations of the letters ABCDEFGH contain

a) the strings CAB and BED?

myans.) 5P5 = 5! = 120 ways

b) the strings BCA and ABF?

myans.) permutation not possible since B cannot be followed by both C and F at the same time.

Thanks for any help!
Well, the letters CAB can appear in 6 positions...

CABxxxxx, xCABxxxx, xxCABxxx, xxxCABxx, xxxxCABx and xxxxxCAB

BED cannot appear in the last 2 positions, so there are only 4 positions that permit CAB and BED.

Of these, the remaining letters have 3! permutations each...

So there are 4*6 = 24 permutations allowing the combination CABED.

3. Originally Posted by robocop_911

Q.1 How many permutations of the letters ABCDEFGH contain

a) the strings CAB and BED?

myans.) 5P5 = 5! = 120 ways

b) the strings BCA and ABF?

myans.) permutation not possible since B cannot be followed by both C and F at the same time.

Thanks for any help!
I take my word back!

4. Originally Posted by sean.1986
Well, the letters CAB can appear in 6 positions...

CABxxxxx, xCABxxxx, xxCABxxx, xxxCABxx, xxxxCABx and xxxxxCAB

BED cannot appear in the last 2 positions, so there are only 4 positions that permit CAB and BED.

Of these, the remaining letters have 3! permutations each...

So there are 4*6 = 24 permutations allowing the combination CABED.

Can anyone please verify whether whatever "sean1986" has said is Correct or Not! Thanks!

5. Hello, robocop_911!

Q.1 How many permutations of the letters ABCDEFGH contain

a) the strings CAB and BED?

my answer: . $_5P_5 \:= \:5! \:= \:120$ ways . . . . not quite

The permutation must contain $\boxed{CABED}$.

Then we have four objects to permute: . $\boxed{CABED}\,,\:\boxed{F}\,,\:\boxed{G}\,,\:\box ed{H}$

And there are: . $_4P_4 \:=\:4! \:=\: 24$ ways.

b) the strings BCA and ABF?

my answer: .The permutation not possible
since B cannot be followed by both C and F at the same time.

I agree!