Permutation seating arrangment 5!.3! What wold be the answer for this
How many ways can 3 boys and 4 girls sit in a row if all the boys are sit together.
Answer listed as $5!\cdot3$!
$4+3 = 7$ what is 5 doing here? someone please break the steps of solving this? 
what does mean by 
$5!\cdot3!$ 
 A: The exclamation point denotes the factorial (here is the Wikipedia page). 

The answer to the question can be found as follows. We have a row of 7 chairs,
$$\Box\quad\Box\quad\Box\quad\Box\quad\Box\quad\Box\quad\Box$$
The boys must form a block of 3 chairs. There are 5 such blocks:
$$\blacksquare\quad\blacksquare\quad\blacksquare\quad\Box\quad\Box\quad\Box\quad\Box$$
$$\Box\quad\blacksquare\quad\blacksquare\quad\blacksquare\quad\Box\quad\Box\quad\Box$$
$$\Box\quad\Box\quad\blacksquare\quad\blacksquare\quad\blacksquare\quad\Box\quad\Box$$
$$\Box\quad\Box\quad\Box\quad\blacksquare\quad\blacksquare\quad\blacksquare\quad\Box$$
$$\Box\quad\Box\quad\Box\quad\Box\quad\blacksquare\quad\blacksquare\quad\blacksquare$$
So, when we tell the boys to sit, they must choose one of the 5 blocks of three chairs, and then choose how to arrange themselves in it. The first boy can choose any of the 3 chairs he wants, the second boy can choose from the 2 remaining chairs, and the last boy is forced to take whatever chair in the block is left. Thus, the number of ways of seating the boys is
$$5\times (3!)=5\times 3\times 2\times 1=30.$$
After the boys are seated, the girls choose how to arrange themselves among the 4 remaining chairs. The first girl can choose any of the 4 chairs (etc.), so there are 
$$4!=4\times 3\times 2\times 1=24$$
ways the girls can be seated once the boys have chosen. Thus, overall there are 
$$30\times 24=720=120\times 6=(5!)\times (3!)$$
ways of seating these people.

As others have said, one can ask that the girls' chairs, and the block of chairs for the boys, be chosen together (in this way, we treat the group of boys as "another girl", for a total of 5 "girls" choosing among 5 "seats"); thus, we can have the boys choose one of the 5 "seats" (in reality, it is their block of chairs), the first girl chooses one of the 4 remaining chairs, the second girl chooses one of the 3 remaining chairs, etc. making for
$$5!=5\times 4\times 3\times 2\times 1=120$$
ways of seating the girls and choosing the block of chairs for the boys. Then, all that remains is for the boys to arrange themselves within the block, which there are
$$3!=3\times 2\times 1=6$$
ways to do, making (again) a total of 
$$(5!)\times (3!)=120\times 6=720$$
ways of arranging them.
A: Treat the group of three boys as an additional girl. That makes five girls, which can be arranged in $5!$ ways. Then you can replace the additional girl by any permutation of the three boys, of which there are $3!$.
A: Treat boys as a group so now you have $4$ girls and $1$ group so the are $5!$ ways to arrange.Now boys can be arranged in $3!$ ways in the group .
So total no. of ways are $5!3!$
A: Consider the fact that there are $3!$ ways to seat the boys together (the first boy can sit in 3 places, the second boy can sit in 2 potential places, and the last boy must sit in the remaining place).
Next we can sit the boys between any of the girls (or at the end), this gives $5$ places for them to be seated (3 places between the girls, and 2 at the ends). However, we also need to consider the number of ways the girls can be arranged, which is $4!$. Therefore, we have:
$$(5\times4!)\times3!=5!\times3!$$
Ways to seat the group of people corresponding to the constraints given by your question.
Hope this helps!
A: Temporarily replace the three boys by a single boy. Then you have $5$ people to seat, the $4$ girls and the boy. There are $5!$ ways to arrange $5$ people in a row, so there are $5!$ ways to seat the $4$ girls and the one boy. Now we’ll magically turn the boy back into $3$ boys sitting side by side. That is, if the arrangement was $$G_1G_2G_3BG_4\;,\tag{1}$$ we now have $$G_1G_2G_3BBBG_4\;,$$ with the $3$ boys occupying a block of $3$ seats where the single boy had his seat. But we can arrange the $3$ boys in any order within their block, and there are $3!$ possible orders, so each of the $5!$ original arrangements corresponds to $3!=6$ arrangements of all $8$ children. For example, $(1)$ corresponds to the following $6$ arrangements:
$$\begin{align*}
&G_1G_2G_3B_1B_2B_3G_4\\
&G_1G_2G_3B_1B_3B_2G_4\\
&G_1G_2G_3B_2B_1B_3G_4\\
&G_1G_2G_3B_2B_3B_1G_4\\
&G_1G_2G_3B_3B_1B_2G_4\\
&G_1G_2G_3B_3B_2B_1G_4
\end{align*}$$
Thus, the final total is $5!\cdot3!=120\cdot6=720$ arrangements.
A: Step $1$, choose a starting point for the boys to sit:  $5$ ways
Step $2$, seat the boys:  $3!$
Step $3$, seat the girls:  $4!$
Result $5\times 4!\times3! = 5!3!$
