A gumball machine contains 7 blue, 5 green, and 4 red gumballs - ident

Total No. of ways to select 3 out of 16 is = 16C3

No. Of ways to select one blue ball = 7C1
No. Of ways to select one green ball = 5C1
No. Of ways to select one red ball = 4C1


So. total no. of ways for favourable outcome = 7*5*4 = 140

So Probability = 140/560 = 1/4

When using simple probability, after choosing the first ball, there are only 15 balls to choose from and one less of the color that has already been chosen and after selecting the second, there are only 14 balls to choose from and one less of the color that has already been chosen and so on. Hence order has to be taken into account.

However while using combinations, the order does not matter as we are selecting one ball from one color only each time.

You did a computational mistake, the 3! from the denominator of 16C3 should go up to the numerator.
So your answer would be correct (1/24) x 6 =1/4.

Is there a way to solve this using \(\frac{16!}{7!*5!*4!}\) as the denominator?

Yes.

Total # of outcomes: \(\frac{16!}{7!*5!*4!}\) (the number of arrangements of the mables).

Favorable outcomes: we need the first three marbles to be BGR in any combination, so 3!. The remaining 13 marbles (6 blue, 4 green, and 3 red) can be arranged in 13!/(6!4!3!).

P = (Favorable)/(Total) = \(\frac{(3!\frac{13!}{6!4!3!})}{(\frac{16!}{7!*5!*4!})}=\frac{1}{4}\).

Hope it's clear.

Conceptual question please :

When we do 7 x 5 x 4 we are assuming that order matters and we will get BGR in every permutation. However, if we were to divide this by 3! what will we get ? (Yes its not a whole number)

Blue = 7; Green = 5 ; Red = 4.

Total = 16.

'3' balls are dispersed => \(^{16}\mathrm{C}_3\)

=> \(^{16}\mathrm{C}_3\) = \(\frac{16*15*14 }{ 3*2*1}\) = 560 ways

The probability that it dispenses one of each color => Probability: \(\frac{Desired }{ Total}\)

=> \(^7\mathrm{C}_1\) = 7
=> \(^5\mathrm{C}_1\) = 5
=> \(^4\mathrm{C}_1\) = 4

=> \(\frac{7 * 5 * 4 }{ 560}\)

=> \(\frac{140}{560}\)

=> \(\frac{1}{4}\)

Answer B

Given: A gumball machine contains 7 blue, 5 green, and 4 red gumballs - identical besides colour.
Asked: If the machine disperses 3 gumballs at random, what is the probability that it dispenses one of each colour.

Favorable ways = 7*5*4
Total ways = 16C3 = 16*15*14/3*2*1 = 16*5*7

Probability = 7*5*4/16*5*7 = 1/4

IMO B

I have one doubt . Can somebody guide me please.

In probability method i understand that (7/16∗5/15∗4/14∗3! ) , we are multiplying the result by 3! because the order is assumed as BGR . But in combination method why we don't multiply it by 3! ?

Thanks in advance.

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.