Question

(\frac{21!}{2!2!3!2!3!2!})

Explanation:

Step1: Simplify factorial denominator

First, list out all factorials in the denominator: $2! \times 2! \times 3! \times 2! \times 3! \times 2!$. Count the repeated terms: there are four $2!$ and two $3!$.
$2! = 2$, $3! = 6$. So the denominator becomes $(2!)^4 \times (3!)^2 = 2^4 \times 6^2 = 16 \times 36 = 576$

Step2: Expand numerator and cancel terms

$21! = 21 \times 20 \times 19 \times \dots \times 4 \times 3! \times 2!$
We can cancel the two $3!$ and four $2!$ from numerator and denominator. After canceling, we have:
$\frac{21!}{(2!)^4(3!)^2} = \frac{21 \times 20 \times 19 \times \dots \times 4}{(2!)^2}$
Since we already used two $2!$ and two $3!$ to cancel, we have two remaining $2!$ in the denominator, so $(2!)^2=4$.
Now calculate the product from 4 to 21:
$21 \times 20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4$
First compute this product:
$21 \times 20 = 420$; $420 \times 19 = 7980$; $7980 \times 18 = 143640$; $143640 \times 17 = 2441880$; $2441880 \times 16 = 39070080$; $39070080 \times 15 = 586051200$; $586051200 \times 14 = 8204716800$; $8204716800 \times 13 = 106661318400$; $106661318400 \times 12 = 1279935820800$; $1279935820800 \times 11 = 14079294028800$; $14079294028800 \times 10 = 140792940288000$; $140792940288000 \times 9 = 1267136462592000$; $1267136462592000 \times 8 = 10137091700736000$; $10137091700736000 \times 7 = 70959641905152000$; $70959641905152000 \times 6 = 425757851430912000$; $425757851430912000 \times 5 = 2128789257154560000$; $2128789257154560000 \times 4 = 8515157028618240000$
Then divide by 4:
$\frac{8515157028618240000}{4} = 2128789257154560000$

Answer:

$2128789257154560000$