if nC4 ,nC5 and nC6 are in arithemetic progression. then find value of n
(Note, nC4=n combination 4)
please show workings
#thanks
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
Please solve in a short way
please solve in a short form
please solve in a short method tricks
To find the value of n for which nC4, nC5, and nC6 are in an arithmetic progression, we can use the formula for combinations (nCr) and the properties of arithmetic progressions.
The formula for nCr is given by:
nCr = n! / (r! * (n-r)!)
Where n! represents the factorial of n, which is the product of all positive integers less than or equal to n.
Let's start by writing the expressions for nC4, nC5, and nC6:
nC4 = n! / (4! * (n-4)!)
nC5 = n! / (5! * (n-5)!)
nC6 = n! / (6! * (n-6)!)
Now, since we know that these combinations form an arithmetic progression, we can write the following equation:
2 * (nC5) = (nC4) + (nC6)
Substituting the expressions we found earlier, we have:
2 * [n! / (5! * (n-5)!)] = [n! / (4! * (n-4)!)] + [n! / (6! * (n-6)!)]
To simplify the equation, we can multiply both sides by 4! * 5! * 6! * (n-5)! * (n-4)!:
2 * n! * (n-4)! = n! * (6! * (n-5)!) + n! * (4! * (n-6)!)
Now we can cancel out the factorials:
2 * (n-4)! = 6! * (n-5)! + 4! * (n-6)!
To further simplify, we can divide both sides by (n-4)!:
2 = 6! / (n-5)! + (4! * (n-6)! / (n-4)!)
We can rewrite 6! as 6 * 5 * 4! and simplify further:
2 = (6 * 5) / (n-5) + (4! * (n-6)! / (n-4)!)
Now, we can multiply through by (n-5) and (n-4) to eliminate the denominators:
2(n-5)(n-4) = 30(n-4) + 4(n-6)!
Expanding and simplifying:
2(n^2 - 9n + 20) = 30n - 120 + 4(n-6)!
2n^2 - 18n + 40 = 30n - 120 + 4(n-6)!
2n^2 - 18n + 40 = 30n - 120 + 4n - 24
2n^2 - 52n + 184 = 0
Now we can solve this quadratic equation using factorization, completing the square, or the quadratic formula:
(n - 4)(2n-46) = 0
n - 4 = 0 or 2n - 46 = 0
n = 4 or n = 23
So, the possible values of n for which nC4, nC5, and nC6 are in an arithmetic progression are n = 4 or n = 23.
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.