# How would you use permutations to find the arrangements possible if a line has all the boys stand together?

Jan 31, 2016

If there are $G$ girls and $B$ boys
and all the boys stand together there are
color(white)("XXX")(G+1)G!B! possible arrangements.

#### Explanation:

If the boys are inserted together in a line of $G$ girls,
there are $\textcolor{red}{\text{("G+1")}}$ different places the boys could be inserted each giving a different arrangement.
(One way to see this is to consider how many girls would be to the left of the group of boys; the choices are $\left\{0 , 1 , 2 , \ldots , G\right\}$ for $\left(G + 1\right)$ different possibilities.

The girls could be arranged in color(blue)(G!) different sequences:

$G$ choices for the first position;
$\textcolor{w h i t e}{\text{XX}} \left(G - 1\right)$ for the second (once the first has been determined)
$\textcolor{w h i t e}{\text{XXXX}} \left(G - 2\right)$ for the third (once the first two have been determined)
$\textcolor{w h i t e}{\text{XXXXXX}} \left(G - 3\right)$ for the fourth...
and so on, until ...
$\textcolor{w h i t e}{\text{XXXXXXXXXXXX}} 2$ for the second last position
$\textcolor{w h i t e}{\text{XXXXXXXXXXXXX}} 1$ for the last position.
For a combination of
color(white)("XXX")G xx (G-1) xx (G-2) xx (G-3) xx ... xx 2 xx1 = G! different permutations.

Similarly the boys could be arranged in
color(white)("XXX")color(green)(B!) different permutations.

So
for each of the $\textcolor{red}{\text{("G+1")}}$ locations where the boys could be inserted in the line of girls
$\textcolor{w h i t e}{\text{XXX}}$there are color(blue)(G!) permutations of girsl,
$\textcolor{w h i t e}{\text{XXX}}$and for each of these joint combinations
$\textcolor{w h i t e}{\text{XXXXXX}}$there are color(green)(B!) permutations of boys.

Giving color(red)("("G+1")")*color(blue)(G!)*color(green)(B!) different permuations.