Bunuel wrote:

If three different integers are selected at random from the integers 1 through 8, what is the probability that the three selected integers can be the side lengths of a triangle?

A. 11/28

B. 27/56

C. 1/2

D. 4/7

E. 5/8

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:The task is to pick a combination of three random integers chosen from 1, 2, 3, 4, 5, 6, 7, and 8. The goal is for the three lengths to form a triangle—a condition that will be satisfied if (and only if) the sum of the two smaller integers is greater than the largest integer.

How many total cases are possible? You are picking three items from a set of eight, and the order in which the numbers are picked doesn’t matter, so the total number of ways to select the integers is (8*7*6)/3! = 56.

As for counting the cases that can actually form a triangle, there’s no simple formula or calculation; you’ll actually have to make a list. (As you’ll see, it doesn’t take a terribly long time to make the list, as long as your approach is organized and you start the work immediately. If you hesitated to make this list because you were afraid that it would take too much time, there’s an important lesson to be learned here.)

One observation that can save a great deal of time here is that all of the integers must be 2 or greater; that is, 1 is not among the acceptable numbers. Why? If a and b are integers and a < b, then a must be less than b by at least 1 unit. Therefore, it would be impossible to have a + 1 > b, which would be required for a triangle.

Alternatively, try a couple of cases using 1. You’re picking the two smallest numbers first and making sure the sum is larger than the largest number. 1+2 is not greater than 3 or any other remaining number in the set. 1+3 is not greater than 4 or any other higher number. And so on.

Here’s a complete list of the cases that do form triangles. Pick the two smallest numbers first. Then find a 3rd number that is larger than the other two but smaller than their sum:

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

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

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

(2, 6, 7) (3, 5, 7) (4, 6, 7)

(2, 7, 8) (3, 6, 7) (4, 6, 8)

(3, 6, 8) (4, 7, 8)

(3, 7, 8)

As you write these out, noticing the pattern might help you to make sure you don’t miss any. In the first column (starting with 2), once you pick 3, there’s only one number (4) that is larger than 2 and 3 but smaller than the sum of 2 and 3. The same is true for the rest of that column: once you pick 2 and 4, only one number (5) will work, and so on.

In the second column (starting with 3), there are two possible “endings” for the 3, 4 pair starting point (5 and 6). The same is true for the pair 3, 5 and for the pair 3, 6. The only reason the pair 3, 7 doesn’t have two possible “endings” is that you run out of numbers (the set only goes up to 8).

In the end, there are 22 combinations of integers that form triangles. Since there are 56 combinations overall, the desired probability is 22/56, or 11/28.

What if you messed up the counting and got a number that isn’t in the answers? Estimate to narrow the choices. The total number of cases is less than 1/2, even if you can’t find the exact number, so guess (A) or (B).

The correct answer is (A).

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