Binomial Expansion and Factorials

$B \div A \div A = \dfrac{\dfrac{B}{A}}{A} = \dfrac{B}{A \times A} = \dfrac{200!}{100! \times 100!} = \dbinom{200}{100}$ , which is necessarily an integer.

Here's how to approach the problem without knowing about binomial coefficients.

Let $v_p (n )$ denote the largest power of $p$ which divides $n$ . This means that $p ^ { v_p (n) } \mid n$ but $p ^ { v_p (n) + 1 } \not \mid n$ .
Then, the problem reduces to showing that for all primes, $v_p (B) \geq 2 v_p (A)$ .

Claim 1: $v_p ( n! ) = \lfloor \frac{ n} { p } \rfloor + \lfloor \frac{ n} { p^2 } \rfloor + \lfloor \frac{ n} { p^3 } \rfloor + \ldots$ .
Claim 2: $\lfloor x + y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor$ .

(Proof of the claims are left to the reader.) Using these claims, we obtain the desired inequality:

$v_p (B) = \lfloor \frac{ 200} { p } \rfloor + \lfloor \frac{ 200} { p^2 } \rfloor + \lfloor \frac{ 200} { p^3 } \rfloor + \ldots \geq 2 \lfloor \frac{ 100} { p } \rfloor + 2 \lfloor \frac{ 100} { p^2 } \rfloor +2 \lfloor \frac{ 100} { p^3 } \rfloor + \ldots = 2 v_p (A ).$

Here, A = 100!

       B = 200!

So , B÷A=200!÷100!

    (B÷A)÷A=200!÷[(100!)(100!)]

                   =200 C 100

It's necessarily integer.

$\frac{B}{A}$ = $\frac{200!}{100!}$ = 101 x 102 x 103 x 104 ... x 197 x 198 x 199 x 200 A = 1 x 2 x ... x 100, the largest value in A, 100, cancels out with the the largest value in $\frac{B}{A}$ , so that means logically all the values in A can cancel out with the values in $\frac{B}{A}$ as the the values being multiplied are simply incremented by one i.e. 2 can easily cancel out with any of the even numbers, 3 can easily cancel out with any of the multiples of three ... 99 can easily cancel out with 198 as that is 99 x 2 etc.

Well B/A = 1 x 2 x ...200/1 x 2 x ...100 which you can think of as evaluating to 101 x 102 x ...200. This is because all the factors from 1-100 in B and A cancel each other out. Now any number from 1-100 can be multiplied by another to get a number from 101-200. This means that 101 x 102 x ..200 contains all the factors from 1-100. So if you do 101 x 102 x ...200 / A - you'll get an integer.

Breaking down the equation...

A = 100

B = 200

~ thus ~

A = 1

B = 2

So...

B ➗ A = 2 ➗1 = 2

And

B ➗ A ➗ A = 2 ➗ 1 ➗ 1 = 2

~ thus ~ it is true and B➗A➗A is still an integer