# Number of divisors of $10!$

Determine the amount of divisors of $$10!$$

This is a question in my combinatorics textbook, so I need to somehow reduce this to an elementary counting problem like combinations, permutations with or without repetition. I just don't see it. I can determine $$10$$ easy divisors, those are all the numbers $$1$$ through $$10$$ themselves. Then we can consider all possible products of these numbers, but then I get that a number can be either in the product , or not. I would get that it is $$2^{10}=1024$$ but the answer says it should be $$270$$.

What is going wrong here?

Note: $$1$$ and $$10!$$ are included

• You just have to combine $$d(n)=\prod_{p\mid n}\left(\nu_p(n)+1\right)$$ with $$\nu_p(n!)=\sum_{k\geq 1}\left\lfloor \frac{n}{p^k}\right\rfloor$$ to get that the answer is $$(5+2+1+1)(3+1+1)(2+1)(1+1)=270.$$ – Jack D'Aurizio Dec 11 '18 at 0:53

You've overcounted quite a bit, because integer factorization is not unique unless we're talking about primes. So for example, you've got $$4 \cdot 10 = 40 = 5 \cdot 8$$ counted twice.

So rather than thinking about numbers between $$1$$ and $$10$$, think of prime powers. A number is a divisor of $$10!$$ if and only if it is of the form

$$n = 2^a \cdot 3^b \cdot 5^c \cdot 7^d$$ for appropriate ranges of $$a, b, c,$$ and $$d$$. I'll leave it to you to figure out why the values of $$a, b, c,$$ and $$d$$ range from $$0$$ to $$8$$, $$4$$, $$2$$ and $$1$$ respectively, giving

$$(8 + 1)(4 + 1)(2 + 1)(1 + 1) = 270$$

in total.

• Easy, after your hint it all became clear, thank you! $10! = 1 \cdot 2 \cdot3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8\cdot 9 \cdot 10= 2 \cdot 3 \cdot 2^2 \cdot 5 \cdot 2 \cdot 3 \cdot 7 \cdot 2^3 \cdot 3^2 \cdot 5 \cdot 2= 2^8 \cdot 3^4 \cdot 5^2 \cdot 7$ – Algebra geek Dec 11 '18 at 0:23
• You're very welcome. – user296602 Dec 11 '18 at 0:24

Hint $$10!=2^{??}\cdot 3^{??}\cdot5^{??}\cdot 7^{??}$$

Now, any divisor must have the same primes with different powers...

The issue with your approach is that you are double and triple counting some divisors.

For example, you counted $$8$$ as $$8$$ but then you also counted it as $$2 \cdot 4$$. Same way, most numbers divisible by $$8$$ are at least double counted.

You counted $$24$$ as the products $$3 \cdot 8, 4 \cdot 6, 2 \cdot 3 \cdot 4$$ and so on...

• Of course, unique prime factorisation. – Algebra geek Dec 11 '18 at 0:20

Hint for a smaller number: consider $$2^3\cdot 3^2$$ = 72. To make one of its factors, how many times do you want to use $$2$$?