Manpreet Singh wrote:Hey Mitch,
Can you explain in little depth. How it becomes a combination problem???/
Let's say that there are only 3 cites: A, B and C.
Every possible distance between these 3 cities requires an entry on the chart:
AB (the distance from A to B)
AC (the distance from A to C)
BC (the distance from B to C)
Total entries = 3.
The 3 entries needed -- AB, AC, and BC -- are all of the distinct COMBINATIONS OF 2 that can be formed from the three letters A, B, and C.
I applied this same reasoning in my solution above.
To determine the number of entries required for 30 cities, we must count the number of distinct COMBINATIONS OF 2 that can formed from 30 options:
(30*29)/(2*1) = 435.
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