Formation of ranks with $m$ girls and $r$ boys

combinatorics

In how many ways can $r$ boys and $m$ girls queue up so that girls are together?

Attempt: If, for example, $r=5$ and $m=3$. By grouping the number of girls together and exchanging them, we would have a total of $3! \cdot 4!$. Analogous to this problem, we would have $m! (r-m)!$, with $r>m$. The answer is $m!(r+1)!$, why am I wrong?


Solution 1:

The no of ways to group $m$ girl is $m!$. The no of ways to group $r$ boys + 1 group of girl is $(r+1)!$. Counting the group of girl as 1 will make sure they are always together.

Therefore the answer is $m!(r+1)!$

Solution 2:

Let the girls together as $G$, and $r$ boys as $b_1, b_2, \cdots b_r$.

Then there are $m!$ possibilities to generate $G$, and there are $(r+1)!$ ways to line up $r+1$ elements, which are $G, b_1, b_2, \cdots b_r$.

So the answer is $m!(r+1)!$

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