1. ## Combinations/ Permutation

i have discovered a question that i am stuck on again. Help would be appreciated!

In how many ways can you arrange the word BOOKKEEPER without having identical letters (such as OO) beside each other?

2. I am going to be brave here and give it a go. Perhaps if I'm overcounting or undercounting, someone will be along.

Start by arranging the R,P, and B with spaces in between and on the ends.

_B_P_R_

This gives us 4 positions to choose 3 and 3! ways to arrange the B,P,R.

After some head-scratching and trying to account for the different cases.

We get $\displaystyle 3!C(4,3)C(7,2)C(9,2)=18144$

3. The trouble with that answer is that it would never count the following:
EOEBOKEPRK.
That is, it may not count cases in which two repeats are separated by another repeater as in the example above.

4. Oh well, this isn't an easy problem.

We arrange the P,R,B in a line with the gaps. This leaves 4 places we can choose 3 to place the E. This gives the 24 ways as per the last post.

Then, we have 6 letters and we arrange them with gaps in between on the ends. This gives 7 positions we can choose 2 to place the K. C(7,2)=21

Now we have 8 letters and we arrange with gaps in between and on the ends. This gives 9 places and we can choose 2 to place the O. C(9,2)=36 ways.

We have 24*21*36=18144.

That seems OK, but if I am missing something then..............

5. I understand how you got the above answer. I also agree that it is a lower bound for the correct answer.

But once we have placed the E’s there is never a way to get a string containing “EOE…” or a string containing “EKE…” because there is never a gap between two E’s (there are not together to begin with). If we place the O’s first then we will never get a string containing “OEO…” or a string containing “OKO…” for the same reason.

I am willing to wager that who ever wrote this question did not think it though. If there were just two E’s, then something like to old couples problem could be used.

6. It always surprises me how the most experienced ‘counter’ can be blinded by a false start.
Suppose that O is the set of rearrangements in which the two O’s are together; K is the set of rearrangements in which the two K’s are together; and E is the set of rearrangements in which the three E’s are together.
We want to count the number $\displaystyle \left( {O^c \cap K^c \cap E^c } \right) = \left( {O \cup K \cup E} \right)^c$.

$\displaystyle \left| O \right| = \left| K \right| = \frac{{9!}}{{\left( {2!} \right)\left( {3!} \right)}}\,\& \,\left| E \right| = \frac{{8!}}{{\left( {2!} \right)^2 }}$.

$\displaystyle \left| {O \cap K} \right| = \frac{{8!}}{{3!}}\,\& \,\left| {O \cap E} \right| = \left| {K \cap E} \right| = \frac{{7!}}{{2!}}$

$\displaystyle \left| {O \cap K \cap E} \right| = 6!$

Can you put that all together?