Question 1206180
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<pre>

If it is assumed that the value of this number remains unchanged, then there is only one arrangement.

It is what you see.


If any permutations are allowed and we do not care about the value of the number, 

then we have 7 positions, in all, and 3 (three) different digits.

The digit 3 is repeated five times.


So, the number of different distinguished arrangements is  

    {{{7!/5!}}} = 6*7 = 42 after reducing the fraction.    <U>ANSWER</U>



It can be easily explained. We can place the digit 7 in any of 7 positions
and we can place the digit 6 in any of 6 remaining positions.
After that we place the "3's" to fill the 5 remaining positions - it produces one arrangement.
Doing and counting this way, we get 7*6 = 42 different distinguishable arrangements.
</pre>

Solved.


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To see many other similar &nbsp;(and different) &nbsp;solved problems, &nbsp;look into 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.