But It's Not A Combination!

We know that C (101,81) is an integer, and since there is no 101 on the denominator, we can divide the numerator by 101 and it will still be an integer

• Note that $\Large \frac { 101! }{ {20!}{ 81! }}$ $=\begin{pmatrix} 101 \\ 81 \end{pmatrix}$ which is the number of ways of choosing $81$ elements from a set of $101$ is an integers.

• $\Large\frac { 100! }{ 20!80! }$ $=\begin{pmatrix} 100 \\ 80 \end{pmatrix}$ is an integer by the same argument.

• $\begin{pmatrix} 101 \\ 81 \end{pmatrix}=\begin{pmatrix} 100 \\ 80 \end{pmatrix} \times$ $\Large \frac { 101 }{ 81 }$ . Since the left hand side is an integer so is the right hand side. Therefore $81$ divides the numerator. Since $\gcd (101,81)=1$ , $81$ divides $\begin{pmatrix} 100 \\ 80 \end{pmatrix}$ .

Therefore $\begin{pmatrix} 100 \\ 80 \end{pmatrix} \times$ $\Large\frac{1}{81}$ is also an integer.

Here is a direct approach : the maximum exponent of $3$ which divides $100!$ is given by $\lfloor \frac{100}{3}\rfloor + \lfloor \frac{100}{3^2}\rfloor + \lfloor \frac{100}{3^3}\rfloor + \lfloor \frac{100}{3^4}\rfloor = 48$ Similarly, the maximum exponent of $3$ that divides $20!$ and $80!$ are $8$ and $36$ respectively. Thus, the integer $\binom{100}{80}$ is divisible by $3^{48-(8+36)}=3^4=81$ . Hence the number $\frac{100!}{20! 81!}= \frac{\binom{100}{80}}{81}$ is an integer.