What is the number of ways seating three gentleman and three ladies in a row such that each gentleman is adjacent to one lady ?
My attempt – My attempt was to first make three groups each containing a gentleman and a lady and then arrange them.
This leads to $ 3!\times2\times2\times2 $ since the number of ways in which groups can be arranged themselves is $3!$ and then members of each of three groups have two possible arrangements .
Then I multiplied this by the number of ways in which such groups can be formed which is $3!\times3!$.
Hence the answer I got was $1728$ , which is no where close to the answer given in the problem which is $360$.
Where am I going wrong ?
Is there a easier method to accomplish this ?
Best Answer
Your method both under and over counts. It undercounts by neglecting patterns like $GGBBGB$, it overcounts by both permuting and then repopulating patterns. If, instead, you skip the permutations (which can all be realized through the later repopulation) you would get $2^3\times 3!\times 3!=288$ which is the correct answer (if you disallow the two patterns $GGBBGB,\;BGBBGG$)
The hard part is to get all the possible patterns. I'll do that here:
By looking at the possible strings,starting with $B=boy$, there are five possible types of arrangements: $$\{BGGBGB,\;BGGBBG,\;BGBGBG,\;BGBGGB,\;BGBBGG\}$$
Starting with $G=girl$ we also have five possible arrangements, namely $$\{GBBGBG,\;GBBGGB,\;GBGBGB,\;GBGBBG,\;GGBBGB\}$$
Each of these can be populated in $3!\times 3!=36$ ways, hence $360$.
To elaborate on the counting:
Case I. say we start with $B$. Then the next has to be $G$.
Ia. $BGG...$. Fourth must be $B$, whence $BGGB$ but then either $\fbox {BGGBBG}$ or $\fbox {BGGBGB}$ work.
Ib. $BGB...$. We could have $BGBB$ in which case we must have $\fbox {BGBBGG}$ or we could have $BGBG$ win which case we could have either $\fbox {BGBGGB}$ or $\fbox{BGBGBG}$.
Case II (starting with $G$) is similar.