There are some boxes labelled 1, 2, 3, 4... and so on up to 2n, where n>6 such that, for k = 1, 2, 3, 4, .... 2n, there are exactly k boxes labelled k. The total number of boxes is N. Now, consider the following two cases for the number of footballs (fi) in the box labelled i: Case 1: fi=i+1, if i is odd; = i + 2 if i is even. Case 2: fi=i+2, if i is odd; = i + 1 if i is even. The difference between the total number of footballs in the N boxes in Case 1 and Case 2 is:

There are some boxes labelled 1, 2, 3, 4... and so on up to 2n, where

1.	There are some boxes labelled 1, 2, 3, 4... and so on up to 2n, where n>6 such that, for k = 1, 2, 3, 4, .... 2n, there are exactly k boxes labelled k. The total number of boxes is N. Now, consider the following two cases for the number of footballs (fi) in the box labelled i: Case 1: fi=i+1, if i is odd; = i + 2 if i is even. Case 2: fi=i+2, if i is odd; = i + 1 if i is even. The difference between the total number of footballs in the N boxes in Case 1 and Case 2 is:
Answer» There are some boxes labelled 1, 2, 3, 4... and so on up to 2n, where $n > 6$ such that, for k = 1, 2, 3, 4, .... 2n, there are exactly k boxes labelled k. The total number of boxes is N. Now, consider the following two cases for the number of footballs ( $f_{i}$ ) in the box labelled i: Case 1: $f_{i} = i + 1$ , if i is odd; = i + 2 if i is even. Case 2: $f_{i} = i + 2$ , if i is odd; = i + 1 if i is even. The difference between the total number of footballs in the N boxes in Case 1 and Case 2 is:

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