Probability help

[mellow helm]

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[mossy idol]

So think about it this way, 2 out of the m drawers we are not going to pick right?

[winged stirrup]

I am not sure but it may be
(mC(m-2))×(m-2)

[mossy idol]

but there are the extra 2 balls to be put

[radiant dew]

so maybe (m-2 chooses 2)?

is those two are left

[mossy idol]

The problem is overcounting the extra 2 balls

[radiant dew]

right, so we need some place for them

[mossy idol]

First we have $m\choose{m-2}$ possibilities of drawers

[gray wasp]

TheVinkler

[radiant dew]

and what about those 2 extra balls

[mossy idol]

That's the problem 😦

[radiant dew]

xD

[mossy idol]

What if we predetremine the 2 extra balls?

There are (m-2)^2 ways to do so

[radiant dew]

why ^2?

[mossy idol]

and then there are left $(m-2)!$ ways to orginze everything else

[gray wasp]

TheVinkler

[mossy idol]

2 balls

and m-2 drawers

you can put them wherever you want

[radiant dew]

aa

[mossy idol]

The process that we are doing is as such:

1. choose 2 drawers that will not be used
2. choose 2 balls to put in any drawers as you wish
3. put the rest of the m-2 balls one per drawer

And so we get

$(m-2)^2\cdot (m-2)!\cdot { m\choose{m-2}}$

Not what i ment

[radiant dew]

2. but what if we get 4 balls in onde drawer?

[mossy idol]

We cant

[radiant dew]

why?

[mossy idol]

.

We are putting 2 random ones and then one per each

and now that I think about it we are also choosing which balls we are puting

$(m-2)^2\cdot (m-2)!\cdot { m\choose{m-2}}\cdot m\cdot (m-1)$

[gray wasp]

TheVinkler

[mossy idol]

$(m-2)^2\cdot m!\cdot { m\choose{m-2}}$

[gray wasp]

TheVinkler

[mossy idol]

I think that's it

maybe

[radiant dew]

ok thank you

$+close$

[gray wasp]

tasakman95

[mossy idol]

+close