Each participant in a certain study was assigned a sequence

Number of permutations of size three that can be taken from sample of size 8 = 8_P_3

= 8! / (8-3)!

= 8*7*6

= 336

minus 36 not given out

= 300

It's clearly mentioned in the stem that "the sequence A, B, C is different from the sequence C, B, A".

When order of a group/selection matters, we are dealing with permutations/arrangements of the group/selection. So, in this case {a, b, c} is different from {c, b, a}.

Not an universal rule, but below might help to distinguish:
The words "Permutation" and "Arrangement" are synonymous and can be used interchangeably (order matters).
The words "Combination", "Group" and "Selection" are synonymous and can be used interchangeably (order does not matters).

Hope it helps.

As the total # of different sequences for your modified question would be 8^3, then the answer indeed would be \(8^3-36=476\).

Since the order of the sequencing matters, the number of ways to choose and arrange 3 letters from a set of 8 letters is 8P3 = 8!/(8-3)! = 8 x 7 x 6 = 336. Since each participant was assigned to a unique sequence, and 36 of all possible sequences were not assigned, there are 336 - 36 = 300 participants.

Answer: C

I am assuming that you are asking about how this problem would be solved if the order of the letters didn't matter, in which case we are talking about combinations of 3 letters chosen out of 8, rather than permutations of 3 letters chosen out of 8. Here are two thoughts on that:

1) If we have already calculated that there are 336 permutations of 3 letters chosen out of 8, then we can calculate the number of combinations of 8 letters chosen out of 8 by dividing the number of permutations (336) by the number of ways of ordering 3 things (3! = 6). Thus, the number of combinations is 336/6 = 56. We would then subtract 36 from the total number of combinations to get 56 - 36 = 20. As we can see, this is answer choice A, so the testmaker has anticipated that we might make the mistake of calculating combinations instead of permutations!

2) You could also use the combinations formula:

\(nCk = \frac{n!}{(n-k)!k!}\)

Since n = 8 and k = 3, this is:

\(8C3 = \frac{8!}{5!3!} = \frac{8*7*6}{3*2*1} = 56\)

Again, we would get answer choice A.

Please let me know if you have any more questions!

Take the task of creating sequences and break it into stages.

Stage 1: Select the first letter of the sequence
There are 8 letters to choose from.
So, we can complete stage 1 in 8 ways

Stage 2: Select the second letter of the sequence
There are 7 REMAINING letters to choose from (since the three letters must be different).
So, we can complete stage 2 in 7 ways

Stage 3: Select the last letter of the sequence
There are 6 REMAINING letters to choose from.
So, we can complete stage 3 in 6 ways

By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus create a sequence) in (8)(7)(6) ways (= 336 ways)
This means we are able to create enough sequences to accommodate 336 participants in the study.
Since 36 of the possible sequences were not assigned, the number of participants = 336 - 36 = 300

Answer: C

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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The difference is simple:
Your referenced formula deals with dividing items into groups. In the letter problem, we're not dividing; we're creating sequences where the order matters. It's about ordering, not dividing.

Thank you both for your answers, I realize now that I misread the question and didn't take into account the word "different" in the question. On that note, if that word wasn't present (i.e. AAA and AAB could be used), would the answer be 476?

The logic I am using for that is:

= 8 * 8 * 8 - 36 = 512 - 36 = 476

Thanks again!

Why is this a permutation question not combination? The order of the 8 letters doesnt matter (ABC and CAB are different)?

How about 8-3! Pls explain

I tried solving it with similar method as 8c5

Sent from my CPH1727 using GMAT Club Forum mobile app

how would this problem be solved differently is ABC is equal to CBA ( and thus not allowed) ?

Hi Bunuel,
How is the concept of grouping xa into x groups of a each items different from this concept? Wanted to understand the same. Over there, you mentioned that with order, the formula should be (xa)!/(a!)^x. Please help understand this?Bunuel