Ashley and Vinnie work on a sales staff with 8 other salespeople. If f

Hi shriramvelamuri,

The 'math' behind this question can be done in a couple of different ways; your approach IS one of them.

Hi Harley1980,

I understand the logic that you used, but by 'combining' Ashley and Vinnie into '1 person', you've changed the parameters of the question (and the 'inputs' into the equations). Since there are 10 people in total (not 9), you have to account for the possible outcomes that would include only 1 of the 2 people (just Ashley or just Vinnie) and that doesn't happen in your calculations. Those options decrease the probability that BOTH Ashley and Vinnie will be chosen, which is why the correct answer is LESS than 1/4.

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

Favourable cases : Selecting Ashley and Vinnie in 1 way, and then selecting 3 more people from remaining 8.
thus , Fav = 1 * 8C3.
Total cases = 10C5
Thus prob : 8C3 / 10C5 = 2/9
Ans C.

Hi shriramvelamuri,

Yes, your approach and explanation are correct.

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

VERITAS PREP OFFICIAL SOLUTION:

This problem includes a mix of combinatorics and probability. The probability that both Ashley and Vinnie will be chosen is: the number of teams on which they're both chosen divided by the number of total possible teams. In these situations it's usually easier to start with the total. With 10 total people and 5 chosen, the number of combinations (order doesn't matter!) can be calculated as:

10!/(5!5!), which factors out to (10)(9)(8)(7)(6)/[(5)(4)(3)(2)]. That ultimately reduces to 252, but remember that you're doing all of this to find a denominator, the number of total teams. So you may want to leave the math undone; as you factor the numerator 6 with the denominator 2 and 3, and then divide the 10 by 5 and the 8 by 4, you're really left with: 2 * 9 * 2 * 7, which you'll find makes an easier denominator. Then for the number of teams that both Vinnie and Ashley are on. The logic here: if those two are on the team, then there are 8 people left to be chosen for 3 remaining spots. So the calculation is: 8!/(3!5!), which reduces to (8)(7)(6)/[(3)(2)], which is 56 (or you can just leave the 8 * 7 undone). That leaves you with a "teams they're on, divided by teams total" calculation of: (8)(7)/[(2)(2)(9)(7)], in which the 7s cancel and the 8 divided by (2 * 2) leaves a 2 in the numerator. That simplifies to 2/9.

update: wrong explanation

update: wrong explanation

Hi Empower,

Can we then safely assume the answer is 2/9.

[2C2*8C3]/[10C5]=2/9

Hi. Can any one suggest how to do this by probability method?

Posted from my mobile device

Posted from my mobile device

Total Outcome = 10C5 (selecting 5 out of 10)
favourable outcome = 8C3 (selecting 3 out of 8 people as 2 are already selected- Ashley & Vinnie)
Probability = favourable outcome / total outcome = 8C3 / 10C5 = 2/9

There are 10C5 = 10!/(5! x 5!) = (10 x 9 x 8 x 7 x 6)/(5 x 4 x 3 x 2) = 2 x 9 x 2 x 7 = 252 ways to choose 5 people from 10. Assuming Ashley and Vinnie are already chosen, there are only 3 spots left for the other 8 people, and the number of ways to choose 3 people from 8 is 8C3 = 8!/(3! x 5!) = (8 x 7 x 6)/(3 x 2) = 56. Therefore, the probability is 56/252 = 8/36 = 2/9.

Answer: C

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.