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30 Jun 2008, 16:07
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5+4+3+2+1 = 15
From A to all other cities except A = 5
From B to all other cities except A and B = 4
From C to all other cities except A, B and C = 3
From D to all other cities except A, B, C and D = 2
From E to all other cities except A, B, C, D, and E = 1
From F to no city because all cases are already covered.
Answer 1.
From A to all other cities except A = 5
From B to all other cities except A and B = 4
From C to all other cities except A, B and C = 3
From D to all other cities except A, B, C and D = 2
From E to all other cities except A, B, C, D, and E = 1
From F to no city because all cases are already covered.
Answer 1.
30 Jun 2008, 17:25
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I think of it this way, where n=total number of cities. If you try to count you will be screwed when they ask you about 120 cities.
If you filled each square with a dot, that would be n*n=n^2 (# rows * # columns).
Now, you don't need a dot for A to A, B to B, and so on. So for each city that is one less dot (n^2-n).
Finally, putting a dot in every square overlaps, you have twice as many as you need because you have A to B and B to A. So you need to divide your answer by 2, giving you:
(n^2-n)/2. In this case (6^2-6)/2=15
If you filled each square with a dot, that would be n*n=n^2 (# rows * # columns).
Now, you don't need a dot for A to A, B to B, and so on. So for each city that is one less dot (n^2-n).
Finally, putting a dot in every square overlaps, you have twice as many as you need because you have A to B and B to A. So you need to divide your answer by 2, giving you:
(n^2-n)/2. In this case (6^2-6)/2=15
