Show that (n−r)(n+r−1 r)(n r)=n(n+r−1 2r)(2r r). In the LHS (n+r−1r) counts the number of ways of selecting r objects from a set of size n where order is not significant and repetitions are allowed. So you have n people you form r teams and select r captains and select (n−r) players. The RHS divides up a team into 2 sets?

varice2r 2022-09-14 Answered
Show that ( n r ) ( n + r 1 r ) ( n r ) = n ( n + r 1 2 r ) ( 2 r r ) .
In the LHS ( n + r 1 r ) counts the number of ways of selecting r objects from a set of size n where order is not significant and repetitions are allowed. So you have n people you form r teams and select r captains and select ( n r ) players.
The RHS divides up a team into 2 sets?
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Answers (2)

Dalton Erickson
Answered 2022-09-15 Author has 10 answers
Let S be a set of n + r 1 elements. Both sides count the number of ways to select two disjoint sets A , B S of size r and possibly an element c S B.
We first observe that ( n r ) ( n r ) = n ( n 1 r ) as both sides count the number of ways to form a team of size r + 1 with a captain out of n people.
Applying the above on the original LHS we get n ( n 1 r ) ( n + r 1 r ) , which corresponds to selecting A ( r out of n + r 1), then B ( r out of the remaining n 1) and c ( n 1 choices in S B and one option of not choosing 2 r).
The RHS argument goes as follows: choose 2 r elements for both A and B, then choose r of them to make B. Then, as before, there are n options of choosing c S B or none at all.

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cubanwongux
Answered 2022-09-16 Author has 4 answers
Dividing the LHS by the RHS and expanding the definition of the choose notation, we have
( n r ) ( n + r 1 ) ! r ! ( n 1 ) ! n ! r ! ( n r ) ! 1 n ( 2 r ) ! ( n r 1 ) ! ( n + r 1 ) ! r ! r ! ( 2 r ) ! .
By cancelling appropriately, it follows that the LHS divided by the RHS is 1, so they're equal.

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