In a class photograph of $9$ students, $4$ students must be in front and $5$ must be at the back.In how many different ways can the students be seated if the $3$ friends Mariam,Katyusha and Nataly want to sit next to each other?
Sorry it's my first time posting here. So what I did was I assumed the $3$ friends can sit either in front or back.I will count the friends as one element. For front: $(1+1)!\cdot 3!=2!\cdot 3!$ For back: $(1+2)!\cdot 3!=3!\cdot 3!$
And I thought I would sum them up but my answer is not correct.What am I doing wrong?
Think as follows:
The three friends Maryan, Katyusha and Nataly will be referred as $M$, $K$ and $N$. Let us glue them together as one block.
Now, there are two possibilities: 1) they sit on the front or 2) they will sit in the back and we will deal with these cases separately.
1) if they sit in the front there is place for one more student there, this student we can choose from the rest in $6$ different wasy since there are $6$ students left when you remove the trio. Now we have to decidehow the trio and this fourth person will sit, they can be permuted in $2!$ and the rest ($5$ students) will have $5!$ different ways to sit down in the back row. This gives $$ 2!\cdot5! $$ different ways to sit
2) when the friend trio sits in the back row we have to choose $2$ more students out of the remaining $6$ which we can done in $\binom{6}{2}$ different ways and these $3$ entities (the trio and the two chosen students) can be permuted in $3!$ different ways finally the rest will be able to sit in the front row and can choose from $4!$ different ways of sitting down givin a total $$ \binom{6}{2}\cdot 3!\cdot4! $$
Summing the results from these two cases gives
$$ 2!\cdot5!+\binom{6}{2}\cdot 3!\cdot4! $$
Now observe that during all this time we considered the trio as one block and never considered that they also will have different ways to sit down among each other. How many? Well, they are $3$ students and these can be permuted in $3!$ different ways, therefore we have to multiply our result with $3!$ giving the final result
$$ 3!\cdot(2!\cdot5!+\binom{6}{2}\cdot 3!\cdot4!). $$