9 girls are to be seated in a row. Of the 9 girls, 6 insist on being seated in consecutive positions. And of these 6, further 3 girls insist on being seated adjacent to each other. In how many distinct ways can the 9 girls be seated ?
My Attempt :-
Ways of selecting 6 girls out of 9 who want to sit together = $\binom{9}{6}$
Now out of these 6 girls selected we have to further select 3 girls who want to sit adjacent to each other which would be possible in $\binom{6}{3}$ ways
Now for the arrangement purposes, I did:- $4! \cdot 4! \cdot 3!$
$4!$ ways in which the block of 6 girls and 3 separate girls can be arranged
another $4!$ ways in which the block of 3 girls and 3 separate girls in the 6 selected girls can be arranged
and finally $3!$ ways in which the three adjacent sitting girls can be arranged
which gives my final result as $\binom{9}{6}\binom{6}{3} \cdot 4! \cdot 4! \cdot 3!$ ways
Please guide me where am I going wrong, and what would be the correct method?
The girls who want to sit together are some particular girls. For example number the girls 1,2,3,4,5,6,7,8,9. Let the six girls who insist on sitting in consecutive positions be 1,2,3,4,5,6 and among them let 1,2,3 be the girls who again want to be seated adjacent to each other. So these are already fixed. You shouldn't select them explicitly. Hence the remaining computation 4! x 4! x 3! is the answer.