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How many arrangements of $A, B, C, D, E, F, G, H$ are there such that $A$ and B are between C and D ?

Attempt : I am trying to solve this problem using simple counting process like first place can be occupy by 6 persons second occupied by 6 and so forth I tried to manipulate remaining one. But it seems confusing and I am missing some terms is what I feel . Please help me to solve this with effective approach.

N. F. Taussig
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maths student
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    Hint: Pick the positions for $A,B,C,D$ simultaneously. Once you know where those four troublesome letters are... what more do you need to do to make sure that $A$ and $B$ are between $C$ and $D$? What more do you need to do for the remaining letters and remaining positions? – JMoravitz May 27 '20 at 12:19
  • Hint: 1) Pick $4$ spots from $8$ for $A,B,C,D$. 2) Assign $C,D$ to the outer $2$ spots you chose. 3) Assign $A,B$ to the inner spots. 4) Assign the other letters to the remaining spots. Then multiply the numbers of choices in the $4$ steps. – Ned May 27 '20 at 12:20
  • direct counting gives $6720$. – Alexey Burdin May 27 '20 at 12:26
  • @JMoravitz please help I am not getting correct answer. – maths student May 27 '20 at 12:35
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    $\binom{8}{4}\cdot 2\cdot 2\cdot 4! = 6720$. Choose the positions of the four troublesome letters. Choose whether $C$ is first and $D$ is last or if $D$ is first and $C$ is last. Choose whether $A$ is second and $B$ is third or if $B$ is second and $A$ is third. Arrange the remaining letters in the remaining positions. – JMoravitz May 27 '20 at 12:57

3 Answers3

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Another way is by "successive placement" (building up a permutation from scratch) taking into account restrictions

Placing one by one the letters C,D then A,B in between C,D then the rest

# of arrangements $ = (1\cdot2)(1\cdot2)(5\cdot6\cdot7\cdot8) = 6720 $

true blue anil
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If you didn't have the constraint about $A,B$ being between $C,D$, then you'd have $8!=40,320$ arrangements. Now consider the positions of $A,B,C,D$ in each of those arrangements. In each arrangement, exactly one of the six pairs $AB$, $AC$, $AD$, $BC$, $BD$, and $CD$ will lie between the other two of $A,B,C,D$. So you can classify the $40,320$ arrangements into $6$ classes according to which pair lies between the others. Each of those $6$ classes has, by symmetry, the same number of elements, $40,320\,/\,6$. And your problem asks for the number of arrangements in one of the classes. So the answer is $40,320\,/\,6=6720$.

Andreas Blass
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This is a Stars and Bars problem in disguise. For Stars and Bars theory, see this article and this article.

Reserve the factors of

  • $2!~$ as number of ways of permuting A,B.

  • $2!~$ as number of ways of permuting C,D.

  • $4!~$ as number of ways of permuting E,F,G,H.

With the above factors reserved, assume that A precedes B, C precedes D, and that E, F, G, H appear in that order. Now, consider the following tableau:

_ C _ A _ B _ D _

The above tableau indicates that $~5~$ islands are created between the letters C, A, B, and D. Reading these islands left to right, let $~x_1, \cdots, x_5,~$ denote the respective sizes of these islands.

Then, the final factor will be the number of solutions to

  • $x_1 + \cdots + x_5 = 4.$

  • $x_1, \cdots, x_5 \in \Bbb{Z_{\geq 0}}.$

By Stars and Bar theory, the final factor is $~\displaystyle \binom{8}{4} = 70.$

So, the final answer is

$$2! \times 2! \times 4! \times 70 = 6720.$$

user2661923
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