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Question

Let a1a2...an=m be positive integers. Denote by bk the number of those ai for which aik. Prove that a1+a2+...an=b1+b2+...+bm.
  1. mj=1jb, if counted by rows.
  2. mj=1bj, if counted by columns.
  3. mj=bj, if counted by columns.
  4. None of these

A
mj=bj, if counted by columns.
B
mj=1bj, if counted by columns.
C
mj=1jb, if counted by rows.
D
None of these
Solution
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Let us consider a n×m array of numbers (xij) defined as follows : in row i, the first ai entries are equal to 1 and the remaining mai entries are equal to 0. For instance, if n=3 and a1=2,a2=4,a3=5, array is
11000
11110
11111
Now , if we examine column j, we notice that the number of 1's in that column equals the number of those ai greater than or equal to j , hence bj. The desired result follows by adding up in two ways the 1's in the array. The total number of 1's is ni=1ai, if counted by rows, and mj=1bj, if counted by columns.

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