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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.)
Re: Each participant in a certain study was assigned a sequence
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27 May 2010, 10:44
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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.
# 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]
27 May 2010, 11:13
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?
Re: Each participant in a certain study was assigned a sequence
[#permalink]
27 May 2010, 11:22
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Expert Reply
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.
# 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 \(8^3-36=476\). _________________
Re: Each participant in a certain study was assigned a sequence
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27 May 2017, 05:43
Expert Reply
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.)
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
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04 Oct 2017, 19:53
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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.
Re: Each participant in a certain study was assigned a sequence
[#permalink]
09 Oct 2017, 15:55
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Expert Reply
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.
Re: Each participant in a certain study was assigned a sequence
[#permalink]
15 Jun 2018, 13:36
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.
Re: Each participant in a certain study was assigned a sequence
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15 Jun 2018, 21:44
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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).
Re: Each participant in a certain study was assigned a sequence
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03 Dec 2018, 16:09
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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!
Re: Each participant in a certain study was assigned a sequence
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10 Jul 2019, 15:57
Order of sequencing matters since A-B-C is different from B-A-C even though the same letters are used. 36 of the patients, who each has a unique code, are taken out of the total codes we calculate. We are told that the 'sequence' is a 3 letter code consisting of DIFFERENT letters. This means we can't repeat letters.
Therefore, since repetitions aren't allowed the total actual # of "sequences" equals 8*7*6 = 56*6=336.
Now 36 of these aren't used. 336-36 = 300. _________________
Each participant in a certain study was assigned a sequence
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Updated on: 17 May 2021, 06:38
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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.)
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.
Re: Each participant in a certain study was assigned a sequence
[#permalink]
04 Aug 2020, 09:22
JeffYin wrote:
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!
Please let me know if you have any more questions!
Every one is talking about permutation here but we know that we can solve questions using both formulas . I got my answer using combinations . Do let me know if I have done /assumed anything wrong . Bunuel
so total elements are 8 and we need to select 3 only . We also know that ABC is different from CBA which means order matters . So lets first select and then arrange .
8C3 = 8*7*6 /3*2*1 = 56 Now as order matters lets arrange all these using fundamental dash method .
56 * 3! = 336 Now remove 36 unused letters and we get our ans as 300 .
gmatclubot
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