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hw1_soln_1_2

# hw1_soln_1_2 - OR360 Problem Set 1 Solutions to Problems 1...

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OR360 Problem Set 1 Solutions to Problems 1 and 2 1. Use the definition of the “choose” notation and compute directly, starting from the right-hand side. Bring the two fractions to a common denominator so they can be added, then simplify: n - 1 r - 1 + n - 1 r = ( n - 1)! ( r - 1)!( n - r )! + ( n - 1)! r !( n - r - 1)! = ( n - 1)! r + ( n - 1)!( n - r ) r !( n - r )! = ( n - 1)! n r !( n - r )! = n ! r !( n - r )! = n r For a combinatorial argument, consider a particular element X in a group of n elements. A subset of r elements out of this group either contains X or it does not, hence the total number of subsets equals the sum of the number of subsets containing X and the number not containing X . In order to choose a subset containing X , we take X into the subset, and choose another r - 1 elements out of the remaining n - 1 elements in the group, so there are ( n - 1 r - 1 ) ways to do this. In order to choose a subset not containing X , we place X aside and choose the r elements out of the remaining n - 1 elements, so there are ( n - 1 r ) ways to do this. 2. There are many different ways to prove this. The easiest to note is that if A, B, C are indepen-
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