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05 Mar 2019, 11:14
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
61% (01:04) correct
39%
(01:28)
wrong
based on 89
sessions
History
Date
Time
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Not Attempted Yet
There are 3 hangers and 5 shirts in a room. In how many ways, can these 5 shirts be arranged without repetition among the three hangers?
A. 60
B. 40
C. 50
D. 30
E. 20
A. 60
B. 40
C. 50
D. 30
E. 20
07 Mar 2019, 05:48
Kudos
Bookmarks
Hello,
Greetings for the day!
This is a very easy question from the topic of P&C, based on Permutations. However, this question does not necessarily require the usage of the nPr formula. It can be solved without using the nPr formula as well, using the fundamental principles of counting.
Let us have a look at both the approaches.
Method 1 – Using the fundamental principle of counting:
The first hanger can be dealt with in 5 different ways – meaning, any of the 5 shirts can be hung on the first hanger. The second hanger can be dealt with in 4 different ways and the third hanger can be dealt with in 3 different ways. (since we cannot put more than one shirt on one hanger – remember repetition is not allowed)
Since we have to deal with all the three hangers one after the other, these events represent a set of inter-connected events (It is similar to saying Deal with hanger 1 and then with hanger 2 and then with hanger 3, to complete the job).
In case of such types of events which are inter-connected, we have to multiply the values obtained for the individual events.
Hence, the number of ways of doing the assigned job = 5 x 4 x 3 = 60.
Method 2 – Using the nPr formula:
Since the question clearly mentions that the arrangement has to be without repetition of shirts, we can use the nPr formula. Here, n = 5 and r = 3.
Therefore, number of ways to arrange 5 shirts on 3 hangers = 5P3
= (5!) / (5-3)!
= (5!) / (2!)
= 120/2
= 60.
You must have realized by now that, whichever approach you take, this question eventually turns out to be a very simple question to analyse and solve.
Hope this helps!
Cheers
Greetings for the day!
This is a very easy question from the topic of P&C, based on Permutations. However, this question does not necessarily require the usage of the nPr formula. It can be solved without using the nPr formula as well, using the fundamental principles of counting.
Let us have a look at both the approaches.
Method 1 – Using the fundamental principle of counting:
The first hanger can be dealt with in 5 different ways – meaning, any of the 5 shirts can be hung on the first hanger. The second hanger can be dealt with in 4 different ways and the third hanger can be dealt with in 3 different ways. (since we cannot put more than one shirt on one hanger – remember repetition is not allowed)
Since we have to deal with all the three hangers one after the other, these events represent a set of inter-connected events (It is similar to saying Deal with hanger 1 and then with hanger 2 and then with hanger 3, to complete the job).
In case of such types of events which are inter-connected, we have to multiply the values obtained for the individual events.
Hence, the number of ways of doing the assigned job = 5 x 4 x 3 = 60.
Method 2 – Using the nPr formula:
Since the question clearly mentions that the arrangement has to be without repetition of shirts, we can use the nPr formula. Here, n = 5 and r = 3.
Therefore, number of ways to arrange 5 shirts on 3 hangers = 5P3
= (5!) / (5-3)!
= (5!) / (2!)
= 120/2
= 60.
You must have realized by now that, whichever approach you take, this question eventually turns out to be a very simple question to analyse and solve.
Hope this helps!
Cheers
27 Aug 2024, 04:12
Kudos
Bookmarks
Automated notice from GMAT Club BumpBot:
A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.
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A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.
This post was generated automatically.
