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Each participant in a certain study was assigned a sequence

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

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27 May 2010, 11:30
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64% (00:51) correct 36% (00:48) wrong based on 487 sessions

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

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27 May 2010, 11:44
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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.)

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

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27 May 2010, 11:42
3
2
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]

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27 May 2010, 12:13
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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Posts: 49251
Re: Each participant in a certain study was assigned a sequence  [#permalink]

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27 May 2010, 12:22
2
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]

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06 Jun 2016, 12:47
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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Re: Each participant in a certain study was assigned a sequence  [#permalink]

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08 Jun 2016, 02:50
2
1
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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27 May 2017, 06:43
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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

1. It is an nPr problem
2. Total number of possibilities is 8P3= 336
3. 36 of them not assigned
4. Therefore number of participants is 336-36=300
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Re: Each participant in a certain study was assigned a sequence  [#permalink]

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04 Oct 2017, 20:53
3
My approach to solving such questions:

We need 3 letter sequence. _ _ _
First place can be occupied by any of the 8 letters.
Since we have to form sequence using different letters, second place can be occupied by remaining 7 letters and third place by remaining 6 letters.
Hence, 8*7*6 = 336
Out of these, 36 arrangements were discarded. Hence, 336 - 36 = 300.

Kudo if it helps
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Re: Each participant in a certain study was assigned a sequence  [#permalink]

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09 Oct 2017, 16:55
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]

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15 Jun 2018, 14:25
Hi, I got this question wrong because I solved as if it were combo problem. Thus, I set my equation up (8!/5!(3!)) - 36.

Would this have been correct had the problem said that order mattered?
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Re: Each participant in a certain study was assigned a sequence  [#permalink]

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15 Jun 2018, 14:36
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.
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Re: Each participant in a certain study was assigned a sequence  [#permalink]

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15 Jun 2018, 14:58
8 * 7 * 6 - 36 = 300
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Re: Each participant in a certain study was assigned a sequence  [#permalink]

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15 Jun 2018, 21:23

I tried solving it with similar method as 8c5

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

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15 Jun 2018, 22:44
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1
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 &nbs [#permalink] 15 Jun 2018, 22:44
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