Let a1≤a2≤...≤an=m be positive integers. Denote by bk the number of those ai for which ai≥k. Prove that a1+a2+...an=b1+b2+...+bm.
m∑j=1jb, if counted by rows.
m∑j=1bj, if counted by columns.
m∑j=bj, if counted by columns.
None of these
A
m∑j=bj, if counted by columns.
B
m∑j=1bj, if counted by columns.
C
m∑j=1jb, if counted by rows.
D
None of these
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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 m−ai entries are equal to 0. For instance, if n=3 and a1=2,a2=4,a3=5, array is
1
1
0
0
0
1
1
1
1
0
1
1
1
1
1
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 n∑i=1ai, if counted by rows, and m∑j=1bj, if counted by columns.
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