Domino Effect (Number Properties Edition 5, page 118)

[mamajonov]

Dear Manhattan,
I am a little bit confused regarding the computing of the domino effect, because the calculation and outcome is different depending on the teachings of Manhattan and Veritas. I would be highly appreciate if you can indicate any official problem on this topic.
On the page of 118 of Number properties, there is following combinatorics and domino effect problem:
"A miniature gumball machine contains 7 blue, 5 green, and 4 red gumballs, which are identical except for their colours. If the machine dispenses three gumballs at random, what is the probability that it dispenses one gumball each colour?
Solution(in short form):
One case is (7blue/16total)*(5green/15total)*(4red/14total)=1/24
Because three desired gumballs can come out in any order, there are 3!=6 different cases. All of the cases must have the same probability. So, we will multiple 6 to (1/24) and the answer is 1/4 "

However, in the Veritas(Combinatorics and Probability) book, they say we do not multiply to 6. Below is the example of them:
"A jar contains 4 black and 3 white balls. If you reach into the jar and pick two balls at the same time, what is the probability that one ball is white while the other is balck "
Solution: (4/7)*(3/6)=2/7 or (3/7)*(4/6)=2/7
Either way is OK and the asnwer is the same, 2/7

Now I want to know which one is correct approach just for being sure. Is there any official problem similar to that one.
Thanks in advance

[RonPurewal]

That's wrong. The two calculations of 2/7 represent different outcomes, so you have to take both of them into account -- resulting in a probability of 4/7.

If that's not clear, imagine that you remove the balls from the jar one at a time. (Obviously this isn't going to make any difference, since, at the end of the day, you're still removing two balls from the jar.)
Then (4/7)(3/6) represents getting a black ball first, and then a white ball. (3/7)(4/6) represents getting a white ball first and then a black one.
These are distinct possibilities, so you have to add the probabilities together. 2/7 + 2/7 = 4/7.

[RonPurewal]

What might be easier, here, is just to calculate the other two possibilities -- two black or two white -- and then subtract those from 1.
The probability of getting two black balls is (4/7)(3/6), or 2/7.
The probability of getting two white balls is (3/7)(2/6), or 1/7.
Aside from those two, the only remaining occurrence is "one of each color". So, the probability of "one of each color" is 1 - 2/7 - 1/7, or 4/7.

[RonPurewal]

On the page of 118 of Number properties, there is combinatorics and domino effect problem. After computing one outcome (1/24), you calculate it to 6, which is the number of similar outcomes, resulting the 1/4.

If you want to ask a question about a problem from the books, you MUST post the ENTIRE QUESTION, with all answer choices (if there are answer choices).

You can't just give a page number and an oblique reference. Not only is that useless for future readers of the forum, but most moderators don't even have the books on hand while posting on the forum.

Thanks.

[RonPurewal]

Is there any official problem similar to that one.
Thanks in advance

Search the internet for a GMAT Prep problem involving the names "Joshua" and "Jose".

[mamajonov]

Dear Ron,

Thank you very much for your detailed answer. Sorry for not inputting the entire question in the original post. I have just provided the entire question in my revised post.
Regards,

[RonPurewal]

Thanks.

I think my explanation is still adequate; please let me know if you still have any questions.

[SharaT711]

Re-opening this with one specific question. Regarding this part of the explanation:

Because three desired gumballs can come out in any order, there are 3!=6 different cases. All of the cases must have the same probability. So, we will multiple 6 to (1/24) and the answer is 1/4

In combinatorics questions, you often figure out the probability or A and B or A or B, and then you have to divide. How do you determine what to divide by?

A similar question is, how many different combinations can the letters in ASSETS be? And after doing the factorial, you divide by 3 (for the 3 S's). Can you explain this logic?

[RonPurewal]

in the situations you're describing, you're not dealing with probabilities anymore; you're dealing with count numbers ("how many?").
it seems that, on some level, you already understand this: your example (at the end of your post) starts with "how many...", and doesn't mention probability. so, for the moment, i'm going to assume that you know these aren't probability situations.

the base fact here is that, if you have N different things, then the number of orders in which you can arrange them is N!. (you have N choices for the first one, (N - 1) choices for the second one, and so on, until you're down to 1 choice for the last one.)

when some of the N objects are identical, though, you're going to get repeated arrangements. specifically, if there are three identical "s"s, then you can arrange those three "s"s in 3! different ways--all of which are going to look exactly the same.
so, if you just do 6! here, you're going to be counting every possible arrangement 6 different times.
hence 6!/3!.

if you had four "a"s, three "b"s, and two "c"s to arrange, that'd be 9!/(4!•3!•2!).