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30 Nov 2011, 18:08
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Combinations/Permutations:
Q1/Q2: I want to Arrange/Select 5 items 2 of which are identical, in 3 places.
Q3/Q4: I want to Arrange/Select 5 items 2 of which are identical, in 7 places.
Any inputs of how to solve this kind of problems?
Q1/Q2: I want to Arrange/Select 5 items 2 of which are identical, in 3 places.
Q3/Q4: I want to Arrange/Select 5 items 2 of which are identical, in 7 places.
Any inputs of how to solve this kind of problems?
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01 Dec 2011, 02:57
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Responding to a pm:
Q1: I want to select 3 items out of 5 items (2 of which are identical).
e.g. AABCD
There are 2 different cases:
Case 1: All distinct letters (e.g. ABC, ABD, ACD, BCD)
4C3 i.e. 4 ways
Case 2: 2 letters same (e.g. AAB, AAC, AAD)
Select the two As and any one of the other 3 in 3C1 = 3 ways.
Total = 4+3 = 7 ways
Q2: Arrange 3 letters in 3 places out of 5 letters, 2 of which are identical:
You have already selected the letters above. You further need to arrange them here:
Case 1: Arrange the 3 distinct letters you chose above in 3! ways. So selection and then arrangement gives you
4 * 3! = 4! = 24 ways (e.g. ABC, ACB, BCA, BAC etc)
Case 2: Arrange the 2 identical and 1 different letter selected above in
3 * 3!/2! = 9 ways (e.g. AAB, ABA, BAA etc)
Total = 33 ways
Total = 4+3 = 7 ways
Q1: I want to select 3 items out of 5 items (2 of which are identical).
e.g. AABCD
There are 2 different cases:
Case 1: All distinct letters (e.g. ABC, ABD, ACD, BCD)
4C3 i.e. 4 ways
Case 2: 2 letters same (e.g. AAB, AAC, AAD)
Select the two As and any one of the other 3 in 3C1 = 3 ways.
Total = 4+3 = 7 ways
Q2: Arrange 3 letters in 3 places out of 5 letters, 2 of which are identical:
You have already selected the letters above. You further need to arrange them here:
Case 1: Arrange the 3 distinct letters you chose above in 3! ways. So selection and then arrangement gives you
4 * 3! = 4! = 24 ways (e.g. ABC, ACB, BCA, BAC etc)
Case 2: Arrange the 2 identical and 1 different letter selected above in
3 * 3!/2! = 9 ways (e.g. AAB, ABA, BAA etc)
Total = 33 ways
Total = 4+3 = 7 ways
01 Dec 2011, 03:08
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SonyGmat
Responding to a pm:
You don't need to select here since you anyway have 5 items and more than 5 places. We only need to arrange 5 objects in 7 places. Let's imagine we have two more identical items V. Of the 7 places, wherever we get a vacant spot, we put a V there.
So we have AABCDVV - 7 items (2 of which are imaginary) and 7 places
In how many ways can we arrange 7 items in 7 places?
7! ways.
Since we have 2 identical pairs, we get total number of arrangements = 7!/2!*2!
For more on imaginary place fillers, see:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/1 ... ts-part-i/
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
