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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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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.)

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.

# 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.
_________________

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.

# 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?

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.

# 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 \(8^3-36=476\).
_________________

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.)

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
_________________

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.

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.

Answer: C
_________________

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