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How many ways can 4 children with their mothers be seated in a column such that a child placed always in front of his mother?

The answer in the book is $$\frac{8!}{2!2!2!2!}$$

I didn't understand how to get this answer, please help me.

Em.
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user373141
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2 Answers2

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With no restrictions, there are $8$ people who need to be arranged, this can be done in $8!$ ways. But consider the first child and mother $(c_1, m_1)$. In half of these $8!$ permutations, the child $c_1$ will be in front and in the other half it will be the other way. So we have $\dfrac{8!}{2}$ permutations where $c_1$ is in front of $m_1$.

Now consider $(c_2,m_2)$. In half of the $\dfrac{8!}{2}$ arrangements, $m_2$ is in front of $c_2$ and in the other half $c_2$ is in front of $m_2$. Thus we have a total of $\dfrac{8!}{(2)(2)}$ permuations in which $c_1$ is in front of $m_1$ and $c_2$ is in front of $m_2$. Continue this way for other mother child pairs.

Anurag A
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  • How did you get $\dfrac{8!}{(2)(2)}$ can you elaborate a bit more please?? – user373141 Apr 01 '17 at 05:30
  • @prayersmith think of it this way: suppose there are $n$ ways to make a line with bunch of people. Given two particular people $a$ and $b$, we want to know in how many of the $n$ line-ups will $a$ be before $b$ ? By symmetry the same question can be asked by switching $a$ and $b$. So in half $=n/2$ line ups, $a$ will be before $b$. – Anurag A Apr 01 '17 at 21:53
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The ordering per family is fixed. So you just need to select the two positions for each of the families.

$$ \require{cancel} \binom {8}{2,2,2,2} = \binom 82 \binom 62 \binom 42 \underset{=1}{\binom 22} = \frac{8!}{2!\cancel{6!}}\frac{\cancel{6!}}{2!\cancel{4!}}\frac{\cancel{4!}}{2!2!} = \frac{8!}{2!2!2!2!}$$

Joffan
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