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# Each participant in a certain study was assigned a sequence

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Intern
Joined: 24 May 2010
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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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Question Stats:

63% (00:50) correct 37% (00:55) wrong based on 234 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.

[Reveal] Spoiler:
I got 476, the answer given was 300
[Reveal] Spoiler: OA

Kudos [?]: 16 [1], given: 1

Manager
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Posts: 176

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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
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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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Math Expert
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Posts: 41891

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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

[Reveal] Spoiler:
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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Kudos [?]: 128901 [4], given: 12183

Intern
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Posts: 13

Kudos [?]: 16 [0], given: 1

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

[Reveal] Spoiler:
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!

Kudos [?]: 16 [0], given: 1

Math Expert
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Posts: 41891

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

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27 May 2010, 12:22
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

[Reveal] Spoiler:
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
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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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27 May 2017, 06:43
Expert's post
Top Contributor
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.

[Reveal] Spoiler:
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
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] 09 Oct 2017, 16:55
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