Question 1119602
.
I will provide here another solution.
Probably, it is that "elegant" solution John mentioned in his post.



<pre>
For each given arrangement from the condition, consider all 6 (six) permutations of the three people James, Esther and John 
inside of the greater arrangement.


In this way, you will get <U>all possible 6! = 1*2*3*4*5*6 = 720 arrangements of 6 people in a row</U>.


Therefore, the number of all arrangements under the problem's question is  {{{720/6}}} = 120.


<U>Answer</U>.  The number of all arrangements under the problem's question is  120.
</pre>



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<U>Comment from student</U>: thank you so much Ikleyn. i so much appreciate. i have liked and recommended your page to my friends. 


pls i still have one more. 


Dada want to change her password which is dada112233 but with same letters and number. In how many ways she can do that? thank you in anticipation.
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<U>My responce</U>.  Although it is not formulated explicitly and directly in the condition, I will assume that


<pre>
    4 letters occupy 4 first positions,  and

    6 digits occupy the last 6 positions.
</pre> 

<pre>
So, we have all distinguishable arrangements of the word "dada" in the first 4 positions and all distinguishable arrangements 

of 6 digits "112233" in the last 6 positions.



The word "dada" has 4 letters in all; of them, there are only 2 distinguishable letters each of the multiplicity 2.

The number of all distinguishable arrangements for letters is  {{{4!/(2!*2!)}}} = {{{24/4}}} = 6 in this case.

    (2!*2!) in the denominator stays to account for repeated "a" and repeated "d" with their multiplicities.



The word "112233" has 6 digits in all; of them, there are only 3 distinguishable digits  each of the multiplicity 2.

The number of all distinguishable arrangements for digits is  {{{6!/(2!*2!*2!)}}} = {{{720/8}}} = 90 in this case.

    (2!*2!*2!) in the denominator stays to account for repeated "1"; repeated "2", and repeated "3" with their multiplicities.



Now, to complete the solution, we need simply multiply 6 by 90.


<U>Answer</U>.  There are 540 different ways to create the password in a way described in the problem.</pre>

See the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =https://www.algebra.com/algebra/homework/Permutations/Arranging-elements-of-sets-containing-undistinguishable-elements.lesson>Arranging elements of sets containing indistinguishable elements</A> 

in this site.



Also, &nbsp;you have this free of charge online textbook in ALGEBRA-II in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-II - YOUR ONLINE TEXTBOOK</A>.


The referred lesson is the part of this online textbook under the topic &nbsp;"<U>Combinatorics: Combinations and permutations</U>". 



Save the link to this textbook together with its description


Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson


into your archive and use when it is needed.



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<pre>
    In the future, when you post your message through the "Thank you/comment" window, following my post, please refer

    to the problem ID number (which is  "1119602"  in this case)

    in order I could identify the problem properly and answer under the appropriate post.  


    Thank you.
</pre>