Each • in the mileage table above represents an entry indicating the distance between a pair of the five cities. If the

[Gmat_mission]

Each • in the mileage table above represents an entry indicating the distance between a pair of the five cities. If the table were extended to represent the distances between all pairs of 30 cities and each distance were to be represented by only one entry, how many entries would the table then have?

(A) 60
(B) 435
(C) 450
(D) 465
(E) 900

[spoiler]OA=B[/spoiler]

Source: Official Guide

[Brent@GMATPrepNow]

Table.png

Each • in the mileage table above represents an entry indicating the distance between a pair of the five cities. If the table were extended to represent the distances between all pairs of 30 cities and each distance were to be represented by only one entry, how many entries would the table then have?

(A) 60
(B) 435
(C) 450
(D) 465
(E) 900

[spoiler]OA=B[/spoiler]

Source: Official Guide

APPROACH #1:
Each entry in the mileage table denotes a distinct pair of cities.
We can determine the total number of distinct pairs of cities by using combinations.
When there are 5 cities, the total number of distinct pairs of cities = 5C2 = 10
Noticed that in the given table there are 10 entries. Perfect!

Likewise, if we have a mileage table consisting of 30 cities, the total number of distinct pairs of cities = 30C2
= (30)(29)/(2)(1)
= 435
Answer: B



APPROACH #2:
Notice that, when there are 5 cities in the mileage table, the number of entries = 1 + 2 + 3 + 4
Likewise, if we have a mileage table consisting of 30 cities, the number of entries = 1 + 2 + 3 + . . . . + 28 + 29

One way to calculate this is to apply the following formula:
The sum of the integers from 1 to n inclusive = (n)(n+1)/2

So, 1+2+...........+28+29 = (29)(29+1)/2
= (29)(30)/2
= (29)(15)
= 435
Answer: B

Cheers,
Brent

[Scott@TargetTestPrep]

Table.png

Each • in the mileage table above represents an entry indicating the distance between a pair of the five cities. If the table were extended to represent the distances between all pairs of 30 cities and each distance were to be represented by only one entry, how many entries would the table then have?

(A) 60
(B) 435
(C) 450
(D) 465
(E) 900

[spoiler]OA=B[/spoiler]

Source: Official Guide

Solution:

This problem can best be solved using combinations.

This problem is similar to one in which 30 sports teams are playing in a tournament in which every team plays every other team exactly once. No team plays itself, obviously, and the order of each pairing doesn't matter. [For example, if Team A plays Team B, the pairing of (Team A vs. Team B) is identical to (Team B vs. Team A)]. We would calculate 30C2, or the number of combinations of 30 items taken 2 at a time.

We can solve this problem in the same way:

30C2 = (30 x 29) / 2! = 15 x 29 = 435

Alternate Solution:

Notice that when there are 5 cities, there are 1 + 2 + 3 + 4 entries.

If there were 6 cities, the last column would have contained 5 entries; therefore there would have been 1 + 2 + 3 + 4 + 5 entries. Notice that the last column always contains one less entry than the number of cities.

Thus, if there were 30 cities, there would have been 1 + 2 + 3 + … + 29 = (29 * 30)/2 = 29 * 15 = 435 entries.

Answer: B