OR360Problem Set 1Solutions to Problems 1 and 21.Use the definition of the “choose” notation and compute directly, starting from the right-handside. Bring the two fractions to a common denominator so they can be added, then simplify:n-1r-1+n-1r=(n-1)!(r-1)!(n-r)!+(n-1)!r!(n-r-1)!=(n-1)!r+ (n-1)!(n-r)r!(n-r)!=(n-1)!nr!(n-r)!=n!r!(n-r)!=nrFor a combinatorial argument, consider a particular elementXin a group ofnelements.Asubset ofrelements out of this group either containsXor it does not, hence the total numberof subsets equals the sum of the number of subsets containingXand the number not containingX. In order to choose a subset containingX, we takeXinto the subset, and choose anotherr-1 elements out of the remainingn-1 elements in the group, so there are(n-1r-1)ways to dothis. In order to choose a subsetnotcontainingX, we placeXaside and choose therelementsout of the remainingn-1 elements, so there are(n-1r)ways to do this.2.There are many different ways to prove this. The easiest to note is that ifA, B, Care indepen-
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