Question 1192086
.
<pre>

Any of 6 given digits in the left-most position, giving 6 options;


any of remaining 5 digits in the next position, giving 5 options;


any of remaining 4 digits in the next position, giving 4 options;


any of remaining 3 digits in the last position, giving 3 options.


The <U>ANSWER</U> is the product of the number of options in each position


    6*5*4*3 = 360 different numbers.
</pre>

Solved.


------------------


See the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =https://www.algebra.com/algebra/homework/Permutations/Special-type-permutations-problems.lesson>Special type permutations problems</A> 


in this site.  &nbsp;Find there many solved similar problems and learn the subject from there.


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.