Permutations with restrictions Three  ladies  have  each  brought  a  child  for  admission  to  a  school. The  head  of  the  school  wishes  to  interview  the  six  people  one  by  one, taking  care  that  no  child  is  interviewed  before  its  mother.  In  how many different  ways  can  the  interviews be  arranged?
I tried to apply concept of balanced parenthesis 
so there can be 5 different permutations for balanced parenthesis with 6 strings
12 $\left \{ \left \{  \right \} \left \{  \right \}  \right \}$
6  $\left \{   \right \} \left \{  \right \} \left \{  \right \}$
12 $\left \{   \right \} \left \{ \left \{  \right \}  \right \} $
12   $ \left \{ \left \{  \right \}  \right \} \left \{   \right \} $
36 $ \left \{ \left \{ \left \{   \right \}   \right \}  \right \} $
i'm getting 78 permutations but answer is 90 
which permutations are missing?
 A: One way to organize the count is to case it by the following pattern (where $\text{P}$ denotes one of the parents, and $\text{C}$ denotes one of the children):


*

*$\text{PPPCCC}:\;\;(3!)(3!)=36\;$ways

*$\text{PPCPCC}:\;\;(3)(2)(2)(1)(2)(1)=24\;$ways

*$\text{PPCCPC}:\;\;(3)(2)(2)(1)(1)(1)=12\;$ways

*$\text{PCPPCC}:\;\;(3)(1)(2)(1)(2)(1)=12\;$ways

*$\text{PCPCPC}:\;\;3!=6\;$ways


which yields a total of
$$36+24+12+12+6=90$$

For an alternate, more sophisticated approach, you can argue as follows . . .

Label the parents-child pairs as$\;\;P_1,C_1,\;\;P_2,C_2,\;\;P_3,C_3$.

Place $P_1,C_1$ in the queue, but allow flexible space before them, after them, and between them, to allows for potential later insertions.

To place $P_2,C_2$, since there are $3$ available flexible spaces, there are ${\large{\binom{3}{1}}}$ ways to place $P_2,C_2$ in the same space, and ${\large{\binom{3}{2}}}$ ways to place $P_2,C_2$ in two distinct spaces.

As before, allow flexible space before the beginning of the queue, after the end of the queue, and between any two people in the queue, to allow for potential later insertions.

To place $P_3,C_3$, since there are $5$ available flexible spaces, there are ${\large{\binom{5}{1}}}$ ways to place $P_3,C_3$ in the same space, and ${\large{\binom{5}{2}}}$ ways to place $P_3,C_3$ in two distinct spaces.

It follows that the total number of ways is
$$
\left(
{\small{\binom{3}{1}}}+{\small{\binom{3}{2}}}
\right)
\left(
{\small{\binom{5}{1}}}+{\small{\binom{5}{2}}}
\right)
=(6)(15)
=90
$$
With the same reasoning, for $n$ parent-child pairs, the number of ways is
\begin{align*}
&\prod_{k=2}^n \left[{\small{\binom{2k-1}{1}}}+{\small{\binom{2k-1}{2}}}\right]\\[4pt]
=\;&\prod_{k=2}^n k(2k-1)\\[4pt]
=\;&\left(\prod_{k=2}^n k\right)\left(\prod_{k=2}^n (2k-1)\right)\\[4pt]
=\;&n!\left(\prod_{k=2}^n (2k-1)\right)\\[4pt]
=\;&n!\left(\frac{(2n-1)!}{2^{n-1}(n-1)!}\right)\\[4pt]
=\;&\frac{n\Bigl((2n-1)!\Bigr)}{2^{n-1}}\\[4pt]
=\;&\frac{(2n)!}{2^{n}}\\[4pt]
\end{align*}
which, for $n=3$, yields
$$\frac{6!}{2^3}=\frac{720}{8}=90$$
matching our previously calculated result.

But there's an even easier way . . .

Start by placing the $2n$ people in the queue in any order, yielding $(2n)!$ ways.

But with that count, each parent-child pair can occur in two ways; once in an acceptable order, and once in an unacceptable order. 

Thus, our count of $(2n)!$ is inflated by a factor of $(2)\cdots(2)$, with $n$ factors of $2$, one for each parent-child pair.

Hence, to correct the count, simply divide by $2^n$, yielding
$$\frac{(2n)!}{2^{n}}$$
$$\text{Voila!}$$
A: I guess i have not correctly interpreted the question but I post the answer according to what I thought
Answer 1: No child is interviewed exactly before his mother
Let $M_1,M_2,M_3$ represent the mothers and $B_1,B_2,B_3$ represent the children. 
Hence by inclusion exclusion principle we need to find the number of ways such that a word is formed from the letters $M_1,M_2,M_3,B_1,B_2,B_3$ with the restriction that the strings $B_1M_1$ or $B_2M_2$ or $B_3M_3$ do not occur in the word.
Let these events be represented by the events $A$(the string $B_1M_1$ occurs),  $B$( the string $B_2M_2$ occurs)  and $C$( the string $B_3M_3$ occurs)  respectively. Hence we need to find by inclusion exclusion principle that $$\text {Answer}= 6!- [n(A)+n(B)+n(C)-n(A\cap B) -n(B\cap C) -n(A\cap C) +n(A\cap B\cap C) ]$$
$$\text {Answer}= 6!- [3.5!-3.4!+3!]=426$$
Answer 2: No child is interviewed at any point before his mother
The probability that the first child is not interviewed before his mother is simply $\frac 12$. The same goes for the second and third one. Now since these events are independent of each other hence the probabilities are multiplied
Hence the answer would be $$\frac 12\cdot \frac 12\cdot \frac 12\cdot (\text {Total permutations of 6 people})=\frac {6!}{8}=90$$
As for generalized answer as quasi suggested for $n$ parent child pairs, using the same reasoning above we get $$\frac12\cdot \frac 12\cdot \frac 12\cdot \frac 12\cdot....(\text {$n$ times})\cdot  (\text {Total permutations of $2n$ people})=\frac {2n!}{2^n}$$
A: Partial answer:
Let us start with one adult, followed by three children. One such example is $-A-a-b-c-$, where $A, B, C$ are the adults, and $a,b,c$ are the children. We now have to place the two adults $B$ and $C$ in the spaces.
There are $3$ ways to choose the adult, and $3!$ ways to order the children. There are also $5 \choose 2$ ways to arrange $B$ and $C$, which gives a total of $180$ — twice the original answer.
Of course, some combinations are not valid, because of the 'no child is interviewed before its mother' rule. For example, this methods counts $AabBcC$ when it breaks the rule. 
Feel free to take this answer as inspiration for your solution.
A: Consider the cases of three mothers followed by three children:
$$3!\cdot 3!=36.$$
Consider the cases of the order $M_1,M_2,M_3$:
$$\begin{align}&M_1M_2*M_3** \Rightarrow 2\cdot 2=4 \\ 
&M_1M_2**M_3* \Rightarrow 1\cdot 2=2 \\
&M_1*M_2M_3**\Rightarrow 1\cdot 2=2 \\
&M_1*M_2*M_3* \Rightarrow 1 \end{align}$$
Since there are $3!$ permutations of mothers, then:
$$3!\cdot (4+2+2+1)=54.$$
Hence: $36+54=90$.
