MAT320 Midterm Paper

than or equal to k. When n random cards are removed from the shuffled pile of k black and n red cards, each removed card is either a black or red card. Then the n random cards removed from the shuffled pile consists of x red cards and y black cards, where x and y are non-negative integers x+y=n. After the n random cards that were removed from the shuffled pile are put into the other pile, the number of red cards in the pile that originally consisted of all red cards is k-n+x, and that pile also contains y black cards. The pile that originally consisted of all black cards now has k-y black cards and n-x red cards.

Since x=n-y, the number of red cards in the pile that originally consisted of all red cards is k-n+(n-y)=k-y, which is equal to the number of black cards in the pile that originally consisted of all black cards, and the number of red cards in the pile that originally consisted of all black cards is n-(n-y)=y, which is equal to the number of black cards in the pile that originally had all red cards. Since k, n, x, and y were arbitrary, the above theorem holds true for any k, n, x, and y.

This can be further be illustrated by set notation and manipulation: Let… R = {R1,R2,R3,…,Rk} = The set of all red cards B = {B1,B2,B3,…,Bk} = The set of all black cards N = {R1,R2,R3,…,Rn | n<k} = The set of n cards removed from the red pile Then…

R-N = {Rn+1,Rn+2,Rn+3,…,Rk} = The pile of red cards after n cards removed BUN = { B1,B2,B3,…,Bk,R1,R2,R3,…,Rn} = The Set of black cards combined with the removed red cards Let… M = {B1,B2,…,By,R1,R2,…,Rx | x<n, y<n, x+y=n} = The set of cards removed from BUN