Expert: Sherry Wallin - 11/29/2010

Question
QUESTION: Find n if
[n!/3!(n-5)!]:[n!/5!(n-7)!]= 1:6

ANSWER: Dhananjay~

Are the (n-5)! and the (n-7)! in the denominators or the numerators? If they are in the numerators you need to put parentheses around(n!/3!). Please clarify.

Math Prof

PLEASE CHECK THE PROBLEM, THE WAY YOU TYPED IT. REGARDLESS OF HOW I INTERPRET THE PROBLEM THERE IS NO NATURAL NUMBER ANSWER SO I AM THINKING IT IS MISTYPED???

---------- FOLLOW-UP ----------

QUESTION: They are in the denominator .
now please help  

Answer
Dhananjay~

[n!/[3!(n-5)!)]/[n!/(5!(n-7)!)]) = 1/6
the n! cancel with each other and (n-5)! = (n-5)(n-6)(n-7)! so

[n!/[3!(n-5)!)]/[n!/(5!(n-7)!)]) = 1/6 =>
[1/(3!(n-5)(n-6)(n-7!))]/[1/5(!(n-7)!) ]= 1/6 the (n-7)! cancel so

[1/(3!(n-5)(n-6))]/(1/5!) = 1/6 =>
3! = 3*2*1 & 5! = 5*4*3*2*1 you are left with 5*4 in the denominator.
[1/((n-5)(n-6))]/[1/(5*4)] = 1/6  divide on the left =>
20/[(n-5)(n-6)] = 1/6  cross multiply =>
6*20 = (n-5)(n-6) =>
120 = n^2 -11n + 30 =>
n^2 -11n -90 = 0

This does not factor in the integers! This is why I asked if you entered the problem with a typo!

Math Prof