Solve nC2 = 100C98 for n, with n, 2, 100, 98 denoted

[jshaziza]

Solve the following equation for n: nC2=100C98

nC2 means n and 2 are denoted, same goes for the 100 and 98 in the next formula.

Thx. for any help.

 

[arthur ohlsten]

I assume C means "combinations"
mCp=m!/[ p![m-p]! ]

nC2 = 100C98
n!/[ 2![n-2]!] = 100!/[ 98!2!]
[n][n-1]/2 = 100[99]/2
n^2-n = 9900
n^2-n-9900=0 factor
[n-100][n+99]=0
n=100 answer
n=-99 superfluous answer
I assume n must be a positive number
Arthur

please check math

 

[jshaziza]

For the n!/[2![n-2]!], I can't figure how you got [n][n-1]/2. Could you please explain in detail how you reached that thx.

 

[Mrspi]

For the n!/[2![n-2]!], I can't figure how you got [n][n-1]/2. Could you please explain in detail how you reached that thx.

You had this:

n! / [2! (n - 2)!]

Think about what n! means. n! = n(n-1)(n -2)(n-3).....(1)
But, (n-2)(n-3).....(1) is precisely (n - 2)!

So, we could think of n! as n(n-1)*(n-2)!

n(n-1)*(n-2)!
---------------
2!*(n-2)!

Divide out (n - 2)! from numerator and denominator. Now we have

n(n - 1)
--------
xx2!

But 2! is 2*1, or just 2.....

n(n - 1)
--------
xx2

 

[jshaziza]

Thx. for your clear explanation, it makes sense to me now.