n(n-1)(n-2)(n-3)(n-4)/5! - n(n-1)(n-2)(n-3)/4! = n(n-1)(n-2)(n-3)(n-4)(n-5)/6! - n(n-1)(n-2)(n-3)(n-4)/5!
place all over a common denominator of 6! and then you have a polynomial of the 6th degree. Try a few easy values with synthetic division (0,1,2,3) and you can reduce the degree.
Tedious, but not hard.
if nC4 ,nC5 and nC6 are in arithemetic progression. then find value of n
(Note, nC4=n combination 4)
please show workings
#thanks
6 answers
then
n!/(5!(n-5)!) - n!/(4!(n-4)!) = n!/(6!(n-6)! - n!/(5!(n-5)!)
divide each term by n! and re-arrange
2/(5!(n-5)!) - 1/(4!(n-4)!) = 1/(6!(n-6)!)
realize that
6!=6x5x4! and (n-4)! = (n-4)(n-5)(n-6)!
5!=5x4! and (n-5)! = (n-5)(n-6)!
multiply each term by 6!(n-4)!
2(6)(n-4) - (6)(5) = (1(n-4)(n-5)
12n - 48 - 30 = n^2 - 9n + 20
n^2 - 21n + 98 = 0
(n - 14)(n-7) = 0
n = 14 or n = 7
n!/(5!(n-5)!) - n!/(4!(n-4)!) = n!/(6!(n-6)! - n!/(5!(n-5)!)
divide each term by n! and re-arrange
2/(5!(n-5)!) - 1/(4!(n-4)!) = 1/(6!(n-6)!)
realize that
6!=6x5x4! and (n-4)! = (n-4)(n-5)(n-6)!
5!=5x4! and (n-5)! = (n-5)(n-6)!
multiply each term by 6!(n-4)!
2(6)(n-4) - (6)(5) = (1(n-4)(n-5)
12n - 48 - 30 = n^2 - 9n + 20
n^2 - 21n + 98 = 0
(n - 14)(n-7) = 0
n = 14 or n = 7
Please solve by short method tricks....
Please solve in a short way
please solve in a short form
please solve in a short method tricks
1 answer