Question 1112123
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Hi could you please help me with this question: 
If no digit can be used more than once, find how many numbers that can be formed from the digits 3,4,5,6,7 that are greater than 700?
Its from the chapter of permutations.
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                For some reasons, I assume that your formulation is UNCOMPLETE.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The <U>complete</U> and <U>correct</U> version, in my view, must be like THIS:


<pre>
If no digit can be used more than once, find how many <U>3-digit</U> numbers {{{highlight(cross(that))}}} can be formed from the digits 3,4,5,6,7 that are greater than 700?
</pre>

The major change in the formulation is the addition of the word "3-digit".
Otherwise the problem goes BEYOND the standard school level.



<U>Solution</U>


<pre>
The digit "7" is just fixed/placed in the most-left position, and we can not use it more.


In the second position ("tens" position) you can place any of four remaining digits 3, 4, 5 or 6, which gives you 4 options.


Then for the last ("units") position you have only three remaining digits, which gives you 3 independent opportunities.


Thus, in all you can form 4*3 = 12 different integer numbers satisfying the given conditions.


<U>Answer</U>.  12 numbers.
</pre>



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<U>comment from student</U>: Hi thank you so much for the reply but the answer says its 252 and the question doesn't mention that its 3 digits. 
--------------



<U>My response</U>:  OK, thank you for your message.


<pre>
It means that my suggestion about 3-digit numbers was wrong.

Very good, let us return to the original formulation.  Forget about my attempts to modify it . . . 


So, we need to consider all 5-digit numbers (that, obviously, are all greater than 700); 
                   then all 4-digit numbers (that, obviously, are all greater than 700); 
          and, finally, all 3-digit numbers greater than 700.


It is not so difficult . . . 


5-digit numbers may have any of 5 given digit in the 1-st (left-most) position;
                         any of remaining 4 given digit in the 2-nd position;
                         any of remaining 3 given digit in the 3-rd position;
                         any of remaining 2 given digit in the 4-th position;  and, finally, 
                         only ONE of remaining 1 given digit in the 5-th position.


In all, there are  5! = 5*4*3*2*1 = 120  5-digit numbers with non-repeating given five digits.


Next


4-digit numbers may have any of 5 given digit in the 1-st (left-most) position;
                         any of remaining 4 given digit in the 2-nd position;
                         any of remaining 3 given digit in the 3-rd position;
                         any of remaining 2 given digit in the 4-th, last position.


In all, there are  5*4*3*2 = 120  4-digit numbers with non-repeating given five digits.


And finally, there are only 12 3-digit numbers greater than 700, as I show it in the very first version of my solution.


So, in all we have 120  5-digit numbers + 120 4-digit numbers and 12 3-digit numbers satisfying given condition, or

    120 + 120 + 12 = 252,  which EXACTLY coincide with your answer.


So, I think, the problem is fully resolved.
</pre>

Again, thank you for your feedback !


Please let me know whether everything is clear to you.
If you still have questions, you can message them to me by the same way.

In this case, please refer to the problem ID number 1112123.


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On Combinations and Permutations see the lessons

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =http://www.algebra.com/algebra/homework/Permutations/Introduction-to-Permutations.lesson>Introduction to Permutations</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =http://www.algebra.com/algebra/homework/Permutations/PROOF-of-the-formula-on-the-number-of-permutations.lesson>PROOF of the formula on the number of Permutations</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =http://www.algebra.com/algebra/homework/Permutations/Problems-on-Permutations.lesson>Problems on Permutations</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =http://www.algebra.com/algebra/homework/Permutations/Introduction-to-Combinations-.lesson>Introduction to Combinations</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =http://www.algebra.com/algebra/homework/Permutations/PROOF-of-the-formula-on-the-number-of-combinations.lesson>PROOF of the formula on the number of Combinations</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =http://www.algebra.com/algebra/homework/Permutations/Problems-on-Combinations.lesson>Problems on Combinations</A>

&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> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =https://www.algebra.com/algebra/homework/Permutations/Persons-sitting-around-a-circular-table.lesson>Persons sitting around a cicular table</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =https://www.algebra.com/algebra/homework/Permutations/Combinatoric-problems-for-entities-other-than-permutations-and-combinations.lesson>Combinatoric problems for entities other than permutations and combinations</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =https://www.algebra.com/algebra/homework/Permutations/OVERVIEW-the-lessons-on-Permutations-and-Combinations.lesson>OVERVIEW of lessons on Permutations and Combinations</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 lessons are 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.