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n l =k n l = n n + n n−1 + n n−2 + ...+ n k+1 + n k . Problem 6 (counting the number of increasing vectors) If the ﬁrst component x1 were to equal n, then there is no possible vector that satisﬁes the inequality x1 < x2 < x3 < . . . < xk constraint. If the ﬁrst component x1 equals n − 1 then again there are no vectors that satisfy the constraint. The ﬁrst largest value that the component x1 can take on and still result in a complete vector satisfying the inequality constraints is when x1 = n − k + 1 For that value of x1 , the other components are determined and are given by x2 = n − k + 2, x3 = n − k + 3, up to the value for xk where xk = n. This assignment provides one vector that satisﬁes the constraints. If x1 = n − k , then we can construct an inequality satisfying vector x by assigning the k − 1 other components x2 , x3 , up to xk by assigning the integers n − k + 1 , n − k + 2 , . . . n − 1 , n to the k − 1 k ways. Continuing if x1 = n − k − 1, then we can components. This can be done in 1 obtain a valid vector x by assign the integers n − k , n − k + 1 , . . . n − 1 , n to the k − 1 other components of x. This can be seen as an equivalent problem to that of specifying two k+1 ways. Continuing blanks from n − (n − k ) + 1 = k + 1 spots and can be done in 2 to decrease the value of the x1 component, we ﬁnally come to the case where we have n locations open for assignment with k assignments to be made (or equivalently n − k blanks n to be assigned) since this can be done in ways. Thus the total number of vectors n−k is given by 1+ k 1 + k+1 2 + k+2 3 + ...+ n−1 n−k−1 + n n−k . Problem 7 (choosing r from n by drawing subsets of size r − 1) Equation 4.1 from the book is given by n r = n−1 r−1 + n−1 r . ...
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