Exponent of 7 in $ \binom {100}{50} $ What is the exponent of 7 in $ \dbinom {100}{50} $? It's a question from my exam today and I am completely unsure how to solve it
It seems easy but any help would be appreciated
 A: $^{100}C_{50} = \frac{100!}{50!.50!}$
Logic: You need to calculate the number of $7's$ in numerator and denominator
Number of $7's $ in $50!$ are $8$.So total number of $7's$ in denominator are : $8+8=16$.
Number of $7's$ in $100!$ are $16$.
So $^{100}C_{50}$ has no powers of $7$ left as all $7's$ get cancelled.
SHORT CUT : 
This is how you solve when you are racing against time.
To find number of $k's$ in $n!$, you need to divide n with $k^i$ for $i=1,2,3..$ and add the integer part of the results.
Example: Here $k = 7$ and $n = 100$.
Step$1:$ $\frac{100}{7} = 14.28$ ~ $14$ 
Step$2:$ $\frac{100}{49} = 2.04$ ~ $2$ 
Step$3:$ $\frac{100}{343} = 0.29$ ~ $0$ 
You can stop once you encounter $0$, as all further results will yield you $0$ as well.So if you notice, the number of $7's$ are $14+2=16$, the same result that we got using our $1st$ method where we counted manually.Similarly do for $50!$ and you can see the result for yourself. 
Hope the answer is clear !
A: You should know the following :
$$\binom{100}{50}=\frac{100\cdot 99\cdots 52\cdot 51}{50\cdot 49\cdots 2\cdot 1}.$$
Let $A=100\cdot 99\cdots 52\cdot 51$, $B=50\cdot 49\cdots 2\cdot 1$. Also, let $n_i(A)$ be the number of the multiples of $7^i$ in the numbers from $51$ to $100$. 
Then, what you want is represented as
$$\sum_{i=1} \left(n_i(A)-n_i(B)\right).$$
A: Since this was in your exam....I don't want to give you the complete work.
Here is the method: How many numbers in 100! are divisible by 7? There is 7,14,21,28,35,42,49,56,63, 70,77,84,91,98
So there are a total $\lfloor \frac{100}{7} \rfloor$ or $14$ of them
How many are divisible by $7^2=49$. Exactly  $\lfloor \frac{100}{49} \rfloor$  or $2$.
Hence 100! will have $7^{14+2} $ or $7^{16} as the power of 7
Similarly for 50!:
$$\lfloor\frac{50}{7}\rfloor + \lfloor\frac{50}{49}\rfloor =8 $$
so 50! will have $7^8$ as the power of $7$.
Finally, in
$$100~~ C 50 = \frac{100!}{50! \times 50!}$$
7 will appear as
$$ \frac{7^{16}}{7^8 \times 7^8} = 7^0 $$
So, 7 will not appear as a factor at all!
