Problem Solving

swerve wrote: ↑

Sat Mar 28, 2020 4:25 pm

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

The OA is C

Source: GMAT Prep

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. For more information about the FCP, watch this free video: https://www.gmatprepnow.com/module/gmat- ... /video/775

You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat ... /video/776

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/permutation-t122873.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent

swerve wrote: ↑

Sat Mar 28, 2020 4:25 pm

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

The OA is C

Source: GMAT Prep

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