Division (Probability)
Total should be 2n/(2!)^n+1 right?
There should be an addition 2! in the denominator cause size of distribution is equal??
8 Replies
@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.Class notes (pls let me know if this concept is incorrect)
n -> p, q, r objects
n! / (p! q! r!)
if p=q=r then a extra 3!
we haven't used it in the question, cause we also permute the draws
groups = (2n)!/ [(2!)^n * n!]
total = groups * n!
When in doubt just use nCq.(n-q)Cp.(n-q-p)Cr...
does that mean the coloured balls are distinct
+solved @SirLancelotDuLac @lim me → ∞
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