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A box contains 2 washers,3 nuts and 4 bolts. Items are drawn from the box at random one at time without replacement.the probability of drawing 2washers first followed by 3 nuts and 4 bolts is
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This is by far the easiest probability sum that u can expect, if u avoid silly mistake on this line→ Items are drawn from the box at random one at time without replacement

$2/9∗1/8∗3/7∗2/6∗1/5∗4/4∗3/3∗2/2∗1/1$

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Can you please explain how you got that?

I calculated and found answer to be $24/105$ as follow:

$Total\space possible \space permutation = 9! / (2!\space 3! \space 4!)$

$Total\space possible\space permuations\space that\space we\space want = 2!\space 3!\space 4!$

So $final\space probabilty = (2!\space 3!\space 4!) / (9! / (2!\space 3! \space 4!)) = 24/105$

Can you suggest where I got wrong.