An easy mnemonic way to distinguish between Permutations and Combinations is that a Permutation is a permanent unique iteration where order matters, such as this situation where A | B | C | D is not the same outcome as C |B | D | A. Whereas, a Combination is an iteration where order does not matter and can be changed for instance if we were to select four candidates for two identical job openings this would be a combination since it would not matter whether a person were selected for opening one or two.

As it pertains to this specific question, we should be able to conceptually determine that there is only one method for selecting exactly A | B | C | D as requested. Then, we can determine that there would be four ways to select the first ball, three for the second ball, two for the third ball, and only one for the fourth ball. Since we are selecting each of those four balls, it is necessary to multiply each of the possible outcomes together to determine the total number of possibilities. Therefore, there would be 4 x 3 x 2 x 1 or 4! = 16 total possibilities and we are exactly seeking one of those choices, so you may be able to determine that the correct answer should be 1/16. However, of course neither 1/16 nor 1/4! are available options, so we just recall our mnemonic to determine that this scenario is a Permutation that would be permanently set without the flexibility to change spots to inform that we would need to select the P formula in choice C rather than the C formula in choice B.

What do you mean by formula 2?

Hi Carcass , I did not get it.
Total ways to arrange a,b,c,d are 4! , and only 1 way to arrange it in this particular order a->b->c->d.
So the answer should be 1/4! right?

Please let me know if I am understanding something wrong.
Thanks

That was the OE solution. Edit the typo

Yes

technically the question in that specific answer is wrong formulated.

Now \(4_P_4\) means \(\frac{n!}{(n-r)!}=\frac{4*3*2*1}{4-4}\)= impossible

n must be < r otherwise, the fraction is undefined

The solution should be proposed as simple fraction.

here we do have a permutation in which the order matter

\(4!=4*3*2*1=24\)

Answer is C or 1/24

here permutation and combination are well explained https://gre.myprepclub.com/forum/gre-pe ... tml#p83002