Re: 回复: Re: Permutations and Combinations are CRAZY HARD.
The ratio of the number f combinations of (2n+2) different objects taken n at a time to the number of combinations of (2n - 2) different objects taken n at a time is 99:7 Find the value of n.
HOW THE EFF DO I DO THAT?!
Ah, Fitzpatrick right?
So we have
[(2n +2) C (n)] / [(2n - 2) C (n)] = 99/7
Expand the combination notation
[(2n + 2)!/(2n + 2 - n)! n!] / [(2n - 2)! / (2n - 2 - n)! n!] = 99/7
To start simplifying on the right side, I will flip the second fraction and multiply.
[(2n + 2)! / (n +2)! n!] x [(n - 2)! n! / (2n - 2)!] = 99/7
Expand the factorials and you'll be able to cancel.
[(2n + 2)(2n +1)(2n)(2n - 1)(2n - 2)! / (n+2)(n + 1)(n)(n - 1)(n - 2)! n!] x [(n - 2)! n! / (2n -2)!] = 99/7
The ending factorials from the first fraction cancel out with the factorials on the second fraction and we are left with.
2(n +1)(2n + 1)(2n)(2n - 1) / (n + 2)(n + 1)(n)(n -1) = 99/7
Reduce further by cancelling terms in the numerator with those in the denominator.
4(2n +1)(2n - 1) / (n + 2)(n - 1) = 99/7
(16n^2 - 4) / (n^2 + n - 2) = 99/7
Cross multiply to get rid of fractions
112n^2 - 28 = 99n^2 + 99n - 198
Get everything on one side set equal to zero.
13n^2 - 99n + 170 = 0
Use the quadratic formula to solve for n.
n = [99 +or- sqrt( 99^2 - 4(13)(170)] / 2(13)
n = [99 +or- sqrt(961)]/26
n = 5 or n = 34/13
Therefore, n=5 'cause its an integer.
