SOLUTION: I dont need answer. I just need i) to understand this problem and need to know how to approach it and ii) where can I find similar problems solved or lessons which will illustrate

Algebra ->  Permutations -> SOLUTION: I dont need answer. I just need i) to understand this problem and need to know how to approach it and ii) where can I find similar problems solved or lessons which will illustrate       Log On


   

Question 1164276: I dont need answer. I just need i) to understand this problem and need to know how to approach it and ii) where can I find similar problems solved or lessons which will illustrate how to approach this problem on this site. I tried quite a bit on this site under permutations/combination, lessons, 'text-book' on this site. I am unable to find any guidence. Could someone please point me where I can I go on this site
Question: 7*nP3 = 6*n+1P3 |solve for n
If I expand the first term as 7[n!/(n-3)!] = 6[(n+1)!/(n+1)!-3]
am I on right path?
Thak you
Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
.

Let's go step by step.


(1)  nP3 = n%21%2F%28n-3%29%21.


     Cancel (n-3)! in both numerator and denominator.  You will get


     nP3 = n*(n-1)*(n-2).


           By the way, many people  write and know that  nP3 = n*(n-1)*(n-2)  without any explanations.

           It is simply  another definition  of the nP3 . . . 


     So, the left part of your original equation is 7*n*(n-1)*(n-2).




(2)  Now about the right side.


     I am totally sure that the correct writing for it is  6*((n+1)P3).


     It is  6*(n+1)*n*(n-1),  by the same arguments.




(3)  So, your equation is


         7*n*(n-1)*(n-2) = 6*(n+1)*n*(n-1).


     Next, you cancel factors n and (n-1) in both sides.  You get then


         7(n-2) = 6*(n+1).

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

At this point I stop, leaving the rest for you.


Happy learning (!)


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

    By the way, yesterday I got a message (a "comment from student") asking for my personalized attention.


    I am not 100% sure that this message came from you, I am only 95% sure, based on the context.


    The matter is that this forum is organized and works in the way, that, when I obtain a message from 
    a student, I DON'T know for sure, from whom this message came.


    I asked the founder and the owner of this forum/site, why it is so, and he explained me that this construction
    was made CONSCIOUSLY by him in accordance with the law  to avoid collisions and conflicts between a "student" and a tutor.


    Very strange and very unexpected, but it is so, it is the FACT and we should work under these restrictions.


    For this particular problem, its ID number is  1164276 .


    For any problem, its  ID  number is the first number,  which you see in your page in the  UPPER  LEFT  corner.


    If you post to me without referring to this  ID  number,  I will not know to whom to answer.


    So, if you want to communicate with me and want to have a TWO-WAYS communication, you SHOULD REFER
    to the problem's ID number.  Then I will know to whom to answer, and will answer directly in the SAME SPOT.


    Regarding your request, I am ready to communicate with you and consult you by the same way inside and in frames of this forum,
    as I did it before.


Now, in the frame of this remaining 5% uncertainty, if that message was not from you, then PLEASE IGNORE this my mail.

But if it was from you, then PLEASE confirm.