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To find the number of factors of factorial n, first I tried small values of $n$,
for factorial 2, then 3, 4, 5, 6, 7, 8, and 9. I found a generalisation 
for all values of $n$.

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First find the prime factorisation of $n!$
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<div class="p">
</div>
<div class="p">
<br />
<table width="100%"><tr><td align="center">
    <math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>n</mi><mo>!</mo><mo>=</mo>
<msubsup><mi>p</mi><mrow><mn>1</mn> </mrow>
<mrow><mi>a</mi></mrow></msubsup>

<msubsup><mi>p</mi><mrow><mn>2</mn> </mrow>
<mrow><mi>b</mi></mrow></msubsup>

<msubsup><mi>p</mi><mrow><mn>3</mn> </mrow>
<mrow><mi>c</mi></mrow></msubsup>
<mo>.</mo><mo>.</mo><mo>.</mo></mrow>
    </math>
</td></tr></table>
<br />

</div>
<div class="p">
where $p_1, p_2, ...$ etc. are the prime factors of $n!$ and $a, b,...$ etc 
are the powers of these prime factors.  
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<b>Number of divisors of $n! =  (a + 1)(b + 1)(c + 1) ... $</b>
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To explain the confusion, I take $9!$ as an example. &lt;br&#62;
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<br />
<table width="100%"><tr><td align="center">
    <math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mn>9</mn><mo>!</mo><mo>=</mo>
<msup><mn>2</mn><mrow><mn>7</mn></mrow>
</msup>
<mi>x</mi>
<msup><mn>3</mn><mrow><mn>4</mn></mrow>
</msup>
<mi>x</mi><mn>5</mn><mn>7</mn></mrow>
    </math>
</td></tr></table>
<br />

</div>
<div class="p">
Replacing $n$ with 9, 
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Number of divisors of $9! =    (7+1)(4+1)(1+1)^2  = 8 x 5 x 4 = 160$
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Therefore, $9!$ has $160$ divisors.
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Done by: Ling Xiang Ning, Allan<br />
School: Raffles Institution<br />
Country: Singapore
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[ <i>Editor's note</i>: To understand the reason for this, think of each divisor being 
expressed as a product of prime factors. The number of possible divisors is found 
by working out the number of ways of choosing the prime factors of the divisor. 
The prime $p_1$ may not occur as a factor of the divisor, or it may occur to 
the power 1 or 2 or 3 or any power up to at most a, hence there are $(a+1)$ 
possibilities for a divisor to contain $p_1$ as a factor. Similarly 
there are $(b+1)$
possibilities for a divisor to contain $p_2$ as a factor and $(c+1)$ 
possibilities for 
a divisor to contain $p_3$   as a factor and so on. We multiply these 
numbers of 
possibilities to find the total number of possibilities.
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<div class="p">
The same problem exactly would arise if the classes in a school were each named 
by a different prime number and the number of students in class $p_1$ is a, 
and there are $b$ students in class $p_2, c$  students in class $p_3$
and so on. At 
any time there may be no students from any of the classes in the school building 
or any number from any of the classes. How many different possibilities 
are there for the number of students in the building? Can you see that 
this is the same as the Factorial Fact problem? If not work out the 
answers in some simple cases, for example imagine 2 classes with 3 
children in one and 4 children in the other.]  
" -&#62;
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