Question 1175241
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I solved part (1) yesterday under this link


<A HREF=https://www.algebra.com/algebra/homework/Permutations/Permutations.faq.question.1175182.html>https://www.algebra.com/algebra/homework/Permutations/Permutations.faq.question.1175182.html</A>


https://www.algebra.com/algebra/homework/Permutations/Permutations.faq.question.1175182.html




But I agree to reproduce it here &nbsp;AGAIN, &nbsp;because it is closely connected with part &nbsp;(2).



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<U>Part (1)</U>



For the rest of &nbsp;6 &nbsp;letters, &nbsp;C, D, E, F, G and H, &nbsp;there are &nbsp;&nbsp;6! = 720 possible permutations.


For each of these &nbsp;720 &nbsp;permutations, &nbsp;we have &nbsp;7 &nbsp;positions to place the blocked &nbsp;AB &nbsp;or &nbsp;BA.


THEREFORE, &nbsp;the full number of possible arrangements, &nbsp;satisfying the posed conditions, is


    &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;6!*2*7 = 2*7! = 10080.      &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<U>ANSWER</U>
</pre>

Solved.



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<U>Part &nbsp;(2)</U>



These permutations are &nbsp;COMPLEMENT to  &nbsp;&nbsp;8! = 40320,


so their number is  &nbsp;&nbsp;40320 - 10080 = 30240.    &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<U>ANSWER</U>


Solved.


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I will not solve here part &nbsp;(3), &nbsp;since I think it is &nbsp;TOOOOOO &nbsp;much for one post.


It is not a way to teach people.



It is the way to work instead of them and to perform their job, &nbsp;what I do not want to do.