vindicated

MGRE reveals the solution to last week's Math Beast Challenge problem, and their official answer is indeed (C), 22. However, their method for solving the problem is quite different from mine, and I have to admit I like the simplicity and elegance of their approach:

The easiest way to answer this question is to determine how many introductions Joseph would have to make if no one already knew each other, and subtract the introductions that have already been made.

To determine how many pairs you can make from 8 people, use the combination formula, which can be most easily explained as:

Everything!_____
Picked!NotPicked!

8!__
2!6!

(8)(7)(6)(5)(4)(3)(2)(1)
(2)(1)(6)(5)(4)(3)(2)(1)

(8)(7)(6)(5)(4)(3)(2)(1)
(2)(1)(6)(5)(4)(3)(2)(1)

28

This, incidentally, is the exact same math we would use if asked, “If there are eight teams in a tournament and every team has to play every other team, how many games will there be?” or “If two associates are to attend a conference and eight associates are available to attend, how many pairs of associates could be selected?” In all the cases, the order of the two selected items does not matter. In our problem, introducing Lea to Juan, for instance, is exactly the same thing as introducing Juan to Lea.

We are told in the problem that 6 pairs already know each other:
Mary – Dave
Mary – Edgar
Dave – Edgar
Edgar – Lea
Edgar – Juan
Edgar – Greg

Subtract these six pairs from the 28 total pairs.

28 – 6 = 22

The correct answer is C.

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