Question

if nC4 ,nC5 and nC6 are in arithemetic progression. then find value of n


(Note, nC4=n combination 4)


please show workings
#thanks

Answer 1

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.

Answer 2

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

Answer 3

Please solve by short method tricks....

Answer 4

Please solve in a short way

Answer 5

please solve in a short form

Answer 6

please solve in a short method tricks