If two of the four expressions x+y, x+5y, x-y, and 5x-y are

[mcmebk]

Hi Ron

Somehow I get confused with the other "four letters to four envelops question", or this one:

The Carson family will purchase three used cars. There are two models of cars available, Model A and Model B, each of which is available in four colors: blue, black, red, and green. How many different combinations of three cars can the Carsons select if all the cars are to be different colors?

8*6*4/3!

in this question, we choose one of the two first (X+Y) or (X-Y), probability is 2/4, then to match it, the probability should be 1/3; but since sequence does not matter, don't we divide 2!
so the result would be (2/4*1/3)/2!?

[RonPurewal]

Hi Ron

Somehow I get confused with the other "four letters to four envelops question", or this one:

The Carson family will purchase three used cars. There are two models of cars available, Model A and Model B, each of which is available in four colors: blue, black, red, and green. How many different combinations of three cars can the Carsons select if all the cars are to be different colors?

8*6*4/3!

in this question, we choose one of the two first (X+Y) or (X-Y), probability is 2/4, then to match it, the probability should be 1/3; but since sequence does not matter, don't we divide 2!
so the result would be (2/4*1/3)/2!?

you may want to go back and review the basics before posting this question.

you're currently trying to divide a probability expression by factorials.
that doesn't ever happen; the only things that you divide by factorials are count expressions (i.e., the answers to "how many?" questions, not the answers to "what's the chance?" questions).

[Yylevin90]

im not sure why after realizing that we have only one way of getting the equation do we not 2! the answer bc their are two ways in which the sequence could occur so the answer will be 1/3 instead of 1/6 pls explain thank you

[jlucero]

I think Stacey explained the math perfectly in the initial post, so let me answer your question by using shorthand. Of the four expressions (abc&d), we want to select two. This would be 4!/2!2! = 6 combinations. Out of those six combinations, only one combination will yield the proper form. Therefore, the answer is 1/6. Notice that this method requires us to count the ONE combination that works out of the SIX combinations of expressions. We've done the reducing in each step, so we don't need to do anything afterwards (like multiply by 2).

combinations that yield the proper form: ab
total combinations: ab, ac, ad, bc, bd, cd
answer: 1/6

If you want to do the reducing afterwards, you could say that there are 4*3 = 12 combinations of expressions you could select, and 2 of those combinations work. But this is simply double counting each of the combinations, by flipping the values around.

combinations that yield the proper form: ab, ba
total combinations: ab, ac, ad, ba, bc, bd, ca, cb, cd, da, db, dc
answer: 2/12 = 1/6

Just because there is more than one way to do something, doesn't double the probability IF YOU CAN ALSO DOUBLE THE PROBABILITIES OF EVERY OTHER COMBINATION. If you have to choose one number b/w 0-9 and one letter b/w a-z, there are 10*26= 260 ways to choose a pair. Even if you choose the letter first and the number afterwards, you aren't changing the pairings, just the order of the pairings.

[SUKUMARK689]

I used the following approach, could anyone suggest if this a good approach?
1. There are 4 different numbers and we need to choose in such a way that their product meets the given criteria. This could be done in 4C2 ways which is 6 ways.
2. The number of desired outcomes is 1. Because only (x+y)*(x-y) is possible.

Hence the answer is 1/6.

[RonPurewal]

Yes.