AKHIL KUMAR TRIPATHI wrote:Q.) 3 DIFFERENT INTEGERS ARE SELECTED AT RANDOM FROM THE INTEGERS 1 TO 8.WHAT IS THE PROBABILITY THAT THE THREE SELECTED INTEGERS CAN BE SIDE OF LENGTHS OF A TRIANGLE.
A)11/28 B)27/56 C)1/2 D)4/7 E)5/8


PLEASE GIVE THE SOLUTION

Dear Akhil,

I'm happy to help , but I'll warn you --- there is NO WAY this problem would appear on the GMAT. It takes way too much detail-management counting --- no truly elegant approach occurs to me.

Prob = (# of successes)/(total # of possibilities)

The denominator is easy --- 8C3 = (8*7*6)/(3*2*1) = 56
We compute that with combinations. See:
https://magoosh.com/gmat/2012/gmat-math- ... binations/

The numerator is tricky.

If the longest side is 8, then all numbers which have a sum of eight or less would be forbidden. I will count the forbidden combinations, and then subtract. Here, I am using a Geometry rule known as the Triangle Inequality -- see:
https://magoosh.com/gmat/2012/facts-abou ... -the-gmat/

List the numbers from lowest to highestest. With 8 as the largest side, we couldn't have
(1, a, 8), where a goes from 2 to 7, six possibilities
(2, b, 8), where b goes from 3 to 6, four possibilities
(3, c, 8), where c = 4 or c = 5, two possibilities.
Now, with 7 as the largest side, we couldn't have
(1, d, 7), where d goes from 2 to 6, five possibilities
(2, e, 7), where e goes from 3 to 5, three possibilities
(3, f, 7), where f = 4, one possibility.
Now, with 6 as the largest side, we couldn't have
(1, g, 6), where g goes from 2 to 5, four possibilities
(2, h, 6), where h = 3 or h = 4, two possibilities
Now, with 5 as the largest side, we couldn't have
(1, j, 5), where j goes from 2 to 4, three possibilities
(2, 3, 5) one possibility
Now, with 4 as the largest side, we couldn't have
(1, 2, 4)
(1, 3, 4)
Now, with 3 as the largest side, we couldn't have
(1, 2, 3)

With 8 as the largest side, 6 + 4 + 2 = 12 forbidden cases.
With 7 as the largest side, 5 + 3 + 1 = 9 forbidden cases.
With 6 as the largest side, 4 + 2 = 6 forbidden cases.
With 5 as the largest side, 3 + 1 = 4 forbidden cases.
With 4 as the largest side, 2 forbidden cases.
With 3 as the largest side, 1 forbidden case.

12 + 9 + 6 + 4 + 2 + 1 = 34 forbidden.

This means, 56 - 34 = 22 are allowed, so the probability of picking an allowed combination is 22/56 = 11/28, answer = [spoiler](A)[/spoiler]

Again, that is way more work than the GMAT will expect you to do during the live test.

Does all this make sense?
Mike

Hi Akhil,

Mike's approach to this question is interesting in that he opted to figure out all of the options that WOULDN'T FIT and subtract that number from the TOTAL number of options. While that approach is the best way to tackle certain questions, it's not necessarily the best/fastest approach for this question.

Here, we're limited by a number of factors:

3 DIFFERENT integers
The numbers 1 - 8, inclusive
The Triangle Inequality Theorem (says that any 2 sides, added together, must be > the third side).

There are clearly going to be combos that FIT and combos that WON'T FIT.

I think it will be easier to find the combos that FIT.

The combination formula is really helpful here: 8c3 = 56 groups of 3 different numbers

We're going to list out the options that FIT. Once you create the first list, the other lists will be a lot more obvious (and will be faster to create)

Those with an 8 (remember that the other two numbers, added together, have to be greater than 8)

872
873
874
875
876

863
864
865

854

Those with a 7 (none of these should have an 8 because we put those in the previous list)

762
763
764
765

753
754

Those with a 6 (none of these should have a 7 or an 8 because we put those in the previous 2 lists)

652
653
654

643

Those with a 5 (no 6, 7 or 8 in these)

542
543

Those with a 4

432

Total options that FIT: 9 + 6 + 4 + 2 + 1 = 22

22/56 = 11/28

While I wouldn't expect this exact question to appear on Test Day, the concepts behind it will.

GMAT assassins aren't born, they're made,
Rich