How to do this using permutation and combination

general
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41

I’ve never heard of the “gap method” before, is that what you linked?

edit: oh, guess it’s just another thing I don’t know about

42

you mean k-1 ?

43

I dont know what this method is called in other countries :sweat_smile: but in india, we called it Gap method-
If there are m men and n women (m > n) and they have to sit in a row in such a way that no two women sit together then total no. of such arrangements = (m+1Cn)*m!

44

Yes

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45

I never really bothered to learn combinatorics or identities for it, so that’s just your advantage :stuck_out_tongue:

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46

Okay :sweat_smile:

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47

I tried deriving a formula for it again (and failed). But this is better than the first attempt :stuck_out_tongue:.
Here is the result:
\forall \ n \ge 3: (k-1)k^{n-1} - \displaystyle\sum_{i \ = \ 2}^{n \ - \ 1}\Bigg({{i}\choose{2}}-i+1\Bigg)(k-1)^{n-i+1}
I can’t explain how, it was a mess.
Can you please suggest some nice websites for combinatorics/NT? I want to improve in math.

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48

I see you are just in high school, We would love to know how you got so good.
Tell us about your journey : )

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49

It’s quite simple - I grinded USACO, then grinded CF, and am now grinding AtCoder.

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