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Each signal that a certain ship can make is comprised of 3 different flags hanging vertically in a particular order. How many unique signals can be made by using 4 different flags?
A. 10
B. 12
C. 20
D. 24
E. 36
PS06380
A. 10
B. 12
C. 20
D. 24
E. 36
PS06380
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I would go for D and for following reasons.
A signal can be made using 3 flags. We have total of 4 , so 3 can be chosen out of 4 in 4C3 = 4C1 = 4 ways
The three flags can be arranged in 3 ! ways to make a unique combination. Hence 4 * 3! = 24 ways.
A signal can be made using 3 flags. We have total of 4 , so 3 can be chosen out of 4 in 4C3 = 4C1 = 4 ways
The three flags can be arranged in 3 ! ways to make a unique combination. Hence 4 * 3! = 24 ways.
Another way to look at it.
This is a permutation problem and hence answer will be 4P3 = 24.
This is a permutation problem and hence answer will be 4P3 = 24.
General Discussion
D
4!
i guess it should not be so simple
.
may be i am wrong.
4!
i guess it should not be so simple
.may be i am wrong.
my ans is D but it's not 4!
let's assume the signal is ABC
so A has 4 cases
B has 3 cases
C has 2 cases
so there are 4x3x2=24
let's assume the signal is ABC
so A has 4 cases
B has 3 cases
C has 2 cases
so there are 4x3x2=24
Posts: 386
Consider a flagpole with 3 vertical slots for flags
As we have to select different flags we can do as follows
topmost flag can be paced in 4 ways
middle flag can be placed in 3 ways
bottom most flag can be placed in 2 ways
(order doesn't matter the selection matters)
Total ways = 4 * 3 * 2 = 24 ways
correct answer - D
As we have to select different flags we can do as follows
topmost flag can be paced in 4 ways
middle flag can be placed in 3 ways
bottom most flag can be placed in 2 ways
(order doesn't matter the selection matters)
Total ways = 4 * 3 * 2 = 24 ways
correct answer - D
ydmuley
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Schools: Haas '20 HBS '20 CBS '20 Wharton '20 Stanford '20 Kellogg '20
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Jcpenny
As the order matters:
4 (First Flag) * 3 (Second Flag) * 2 (Third Flag)
24
Hence, Answer is D
Posts: 6
Can anyone please share any theory topic related to this problem I am not getting it
TIA
TIA
ydmuley
Retired Moderator
Joined: 19 Mar 2014
Last visit: 01 Dec 2019
Posts: 807
Location: India
Concentration: Finance, Entrepreneurship
Schools: Haas '20 HBS '20 CBS '20 Wharton '20 Stanford '20 Kellogg '20
GPA: 3.5
Schools: Haas '20 HBS '20 CBS '20 Wharton '20 Stanford '20 Kellogg '20
Posts: 807
Business Analyst
Here you will find all the theory topics explained very well.
https://gmatclub.com/forum/gmat-math-bo ... 30609.html
Posts: 237
One signal has 3 different flags. No. of unique signals that can be made = 3*2*1 = 6.
No. of ways of choosing 3 flags from 4 flags = 4c3 = 4.
No. of signals that can be made = 24. Ans - D.
No. of ways of choosing 3 flags from 4 flags = 4c3 = 4.
No. of signals that can be made = 24. Ans - D.
JeffTargetTestPrep
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Last visit: 05 Jan 2024
Posts: 2,973
Status:Head GMAT Instructor
Affiliations: Target Test Prep
Expert reply
Posts: 2,973
Jcpenny
The number of unique signals is 4P3 = 4 x 3 x 2 = 24.
Answer: D
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Posts: 10,057
Flags available : 4
Signal has: 3 flags
For first flag: 4 choices.
Second flag:3 choices
Third flag: 2 choices
Overall: 4 * 3 * 2 = 24 ways
Answer D
Signal has: 3 flags
For first flag: 4 choices.
Second flag:3 choices
Third flag: 2 choices
Overall: 4 * 3 * 2 = 24 ways
Answer D
A (n,m) = n!/ (n-m)!
A (4, 3) = 4!/ (4-3) ! = 4! = 24
Answer is D
A (4, 3) = 4!/ (4-3) ! = 4! = 24
Answer is D
Combinations + Permutations.
4C3 x 3! = 4 x 6 = 24.
Answer is D.
4C3 x 3! = 4 x 6 = 24.
Answer is D.
Posts: 385
Straightfoward permutation:
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