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Each participant in a certain study was assigned a sequence

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Q49  V42
Re: Each participant in a certain study was assigned a sequence [#permalink]
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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
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Re: Each participant in a certain study was assigned a sequence [#permalink]
Bunuel wrote:
nphilli1 wrote:
Each participant in a certain study was assigned a sequence of 3 different letters from the set {A, B, C, D, E, F, G, H}. If no sequence was assigned to more than one participant and if 36 of the possible sequences were not assigned, what was the number of participants in the study? (Note, for example, that the sequence A, B, C is different from the sequence C, B, A.)

I was hoping someone could walk through the logic on here because I got a different answer than than the answer given in the answer guide.

Thanks

I got 476, the answer given was 300

# of letters = 8, # of ways to choose 3 letters out of 8 when order matters is $$P^3_8=\frac{8!}{(8-3)!}=336$$, which means that there can be 336 different sequences. As 36 sequences were not assigned then 336-36=300 sequences were assigned, so there are 300 participants.

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!
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Re: Each participant in a certain study was assigned a sequence [#permalink]
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nphilli1 wrote:
Bunuel wrote:
nphilli1 wrote:
Each participant in a certain study was assigned a sequence of 3 different letters from the set {A, B, C, D, E, F, G, H}. If no sequence was assigned to more than one participant and if 36 of the possible sequences were not assigned, what was the number of participants in the study? (Note, for example, that the sequence A, B, C is different from the sequence C, B, A.)

I was hoping someone could walk through the logic on here because I got a different answer than than the answer given in the answer guide.

Thanks

I got 476, the answer given was 300

# of letters = 8, # of ways to choose 3 letters out of 8 when order matters is $$P^3_8=\frac{8!}{(8-3)!}=336$$, which means that there can be 336 different sequences. As 36 sequences were not assigned then 336-36=300 sequences were assigned, so there are 300 participants.

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!

As the total # of different sequences for your modified question would be 8^3, then the answer indeed would be $$8^3-36=476$$.
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Re: Each participant in a certain study was assigned a sequence [#permalink]
Why is this a permutation question not combination? The order of the 8 letters doesnt matter (ABC and CAB are different)?
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rjivani wrote:
Why is this a permutation question not combination? The order of the 8 letters doesnt matter (ABC and CAB are different)?

It's clearly mentioned in the stem that "the sequence A, B, C is different from the sequence C, B, A".
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Re: Each participant in a certain study was assigned a sequence [#permalink]
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nphilli1 wrote:
Each participant in a certain study was assigned a sequence of 3 different letters from the set {A, B, C, D, E, F, G, H}. If no sequence was assigned to more than one participant and if 36 of the possible sequences were not assigned, what was the number of participants in the study? (Note, for example, that the sequence A, B, C is different from the sequence C, B, A.)

A 20
B 92
C 300
D 372
E 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.

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Re: Each participant in a certain study was assigned a sequence [#permalink]

I tried solving it with similar method as 8c5

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norovers wrote:
Bunuel can you clarify? Does order mattering mean that a,b,c can also be ordered a different way, such as a,c,b? I always thought that the detailed scenario would be considered an order didn't matter because it could be included in many different ways.

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.
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Re: Each participant in a certain study was assigned a sequence [#permalink]
how would this problem be solved differently is ABC is equal to CBA ( and thus not allowed) ?
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Re: Each participant in a certain study was assigned a sequence [#permalink]
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Anthonydh wrote:
how would this problem be solved differently is ABC is equal to CBA ( and thus not allowed) ?

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!
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Re: Each participant in a certain study was assigned a sequence [#permalink]
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nphilli1 wrote:
Each participant in a certain study was assigned a sequence of 3 different letters from the set {A, B, C, D, E, F, G, H}. If no sequence was assigned to more than one participant and if 36 of the possible sequences were not assigned, what was the number of participants in the study? (Note, for example, that the sequence A, B, C is different from the sequence C, B, A.)

A 20
B 92
C 300
D 372
E 476

*Question edited: 09/10/2016 and OA added.

I got 476, the answer given was 300

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

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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Originally posted by BrentGMATPrepNow on 29 Mar 2020, 07:47.
Last edited by BrentGMATPrepNow on 17 May 2021, 07:38, edited 1 time in total.
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Re: Each participant in a certain study was assigned a sequence [#permalink]
Bunuel wrote:
nphilli1 wrote:
Each participant in a certain study was assigned a sequence of 3 different letters from the set {A, B, C, D, E, F, G, H}. If no sequence was assigned to more than one participant and if 36 of the possible sequences were not assigned, what was the number of participants in the study? (Note, for example, that the sequence A, B, C is different from the sequence C, B, A.)

I was hoping someone could walk through the logic on here because I got a different answer than than the answer given in the answer guide.

Thanks

I got 476, the answer given was 300

# of letters = 8, # of ways to choose 3 letters out of 8 when order matters is $$P^3_8=\frac{8!}{(8-3)!}=336$$, which means that there can be 336 different sequences. As 36 sequences were not assigned then 336-36=300 sequences were assigned, so there are 300 participants.

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
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Re: Each participant in a certain study was assigned a sequence [#permalink]
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Niksthebest wrote:
Bunuel wrote:
nphilli1 wrote:
Each participant in a certain study was assigned a sequence of 3 different letters from the set {A, B, C, D, E, F, G, H}. If no sequence was assigned to more than one participant and if 36 of the possible sequences were not assigned, what was the number of participants in the study? (Note, for example, that the sequence A, B, C is different from the sequence C, B, A.)

I was hoping someone could walk through the logic on here because I got a different answer than than the answer given in the answer guide.

Thanks

I got 476, the answer given was 300

# of letters = 8, # of ways to choose 3 letters out of 8 when order matters is $$P^3_8=\frac{8!}{(8-3)!}=336$$, which means that there can be 336 different sequences. As 36 sequences were not assigned then 336-36=300 sequences were assigned, so there are 300 participants.

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

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.
Re: Each participant in a certain study was assigned a sequence [#permalink]
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