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Consecutive Integers | OG 12

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Consecutive Integers | OG 12

by [email protected] » Sun Sep 16, 2012 2:29 pm
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(Refer to page 168 of OG 12, Q 116 in case the image is not clear.)The image is showing errors here.


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

I was able to do the question by finding 29! and that took time. Did not like the OG Explanation. Rather It was confusing. New ideas to solve this please.
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by GMATGuruNY » Sun Sep 16, 2012 3:25 pm
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Each dot 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 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
Every PAIR of cities must be represented by a dot.
To determine the number of dots that are needed, we simply need to count how many PAIRS can be formed from the 30 cities.
The number of combinations of 2 that can be formed from 30 options = 30C2 = (30*29)/(2*1) = 435.

The correct answer is B.
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by Manpreet Singh » Wed Apr 24, 2013 8:23 am
Hey Mitch,

Can you explain in little depth. How it becomes a combination problem???/
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by GMATGuruNY » Wed Apr 24, 2013 11:34 am
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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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

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