#1**+1 **

**What choose 2 is 120?**

\(\begin{array}{|rcll|} \hline \dbinom{n}{2} &=& 120 \quad & | \quad \dbinom{n}{2} = \dfrac{n}{2}\times \dfrac{n-1}{1} \\\\ \dfrac{n}{2}\times \dfrac{n-1}{1} &=& 120 \\\\ (n)(n-1) &=& 240 \\\\ n^2-n &=& 240 \\\\ n^2-n -240 &=& 0 \\\\ n &=& \dfrac{1\pm \sqrt{1-4(-240)} }{2} \\\\ &=& \dfrac{1\pm \sqrt{1+960} }{2} \\\\ &=& \dfrac{1\pm \sqrt{961} }{2} \\\\ &=& \dfrac{1\pm 31} {2} \quad & | \quad n > 0 \\\\ n &=& \dfrac{1+ 31} {2} \\\\ &=& \dfrac{32} {2} \\\\ &=& 16 \\ \hline \end{array}\)

\(\mathbf{\dbinom{16}{2} = 120}\)

heureka
Mar 8, 2018

#3**0 **

CPhill: Would heureka's method work on large numbers such as: nC4 =3,921,225, where n = 100 ???.

Guest Mar 8, 2018

#4**0 **

I don't think heureka's method will work on very large numbers such as: nC13=153,121,753,939,078,375

There is, however, a method that will work on any n and it goes as follows:

Take the above answer=153,121,753,939,078,375 x 13! =953,492,346,711,122,973,955,200,000

Now, take this huge last number and take the 13th root of it as follows:

953,492,346,711,122,973,955,200,000^(1/13) =118.941. round off this to 119. Then add 13 to it, or

119 + 13 = 132. Now take the average of [119 + 132] / 2 =125.5. Discard the fractional part and keep the integer part or 125, which is answer to n above.

Note: This method will work on ANY n, no matter the size.

Guest Mar 8, 2018