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Yet another lovely perms and combs (1 Viewer)

mrbunton

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this is like placing 41 cards in between groups created/separated by the 9 cards chosen; where in the 8 groups surrounded by at least 2 cards must contain at least one item. So that's like already placing 8 of the cards; making a remainder of 33 cards which are required to be placed inside 10 groups which is (33+9)C9= 42C9 (stars and bars method )

(the cards in this are not considered as distinct but rather it is their order(or better yet the amount of cards before it) that gives them the number (for the 9 chosen cards))


another elegant problem is if u have a pair of red,green and black shoes how much ways can u arrange them so that the left shoe is somewhere left of the corresponding right shoe for each of the pairs, if they are organised into a line. (from amc 2018)
 
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