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in how many ways can 10! be written as the product of two natural number?

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Lets try to prime factorize $10!$ because each if $p^n$ is a factor of a number $x$ for any prime number $p$, then $x$ has at least $p+1$ factors -- $p^0, p^1, p^2, \ldots, p^n.$

$10!=2\times3\times4\times5\times6\times7\times8\times9\times10.$

$=2^8 \times 3^4 \times 5^2 \times 7$

Now, we can count the no. of factors of $10!$ which will be $(8+1) \times (4+1) \times (2+1) \times (1+1) =270.$ For each of the factor $d$ of any number $x$, we can get $x$ by multiplying $d$ by $x/d$ where $x/d$ is another factor of $x$. So, number of ways we can choose $(a,b)$ such that $a\times b = 10!$ will be $\frac{270}{2} = 135.$ In multiplication there is no need to consider $a \times b$ different from $b \times a$, otherwise answer would be $270$.
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 We know that    10! = 3628800 

Prime Factors for the number 3628800 = (28 × 34 × 52 × 7) or(2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 × 7) = 269 factor

 If a Number has ‘N’ factors, Where N is odd (for non square number), Total number of ways it can be expressed as product of 2 natural no’s:

(N+1)/2 ------ If repetition of numbers is allowed. 
(N-1)/2------- If both the no’s has to be unique.   

 ways can 10! be written as the product of two natural number = Total factor / 2 =  269+1  / 2 =   135      ans

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