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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

Micah
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user34304
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3 Answers3

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$^{100}C_{50} = \frac{100!}{50!.50!}$

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 !

lsp
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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 $$\sum_{i=1} \left(n_i(A)-n_i(B)\right).$$

mathlove
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  • Easier: For number $k$ in the denominator that is divisible by $7$, there is correspondingly the number $2k$ in the numerator (and vice versa). So there is no factor of $7$ remaining. – Ted Shifrin Dec 23 '13 at 05:30
  • Oh, I agree with you. Nice! – mathlove Dec 23 '13 at 05:31
  • Ted, the numbers $k=7$, $14$, and $21$ are in the denominator, but $2k$ is not in the numerator for any of them. The pairing-up that works is $7j$ with $7j+49$ for $1\le j \le 7$. – Steve Kass Dec 23 '13 at 05:59
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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!

user44197
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