Why cannot we solve the question by the following steps :
out of 6 places, first wefill all the X's. For that we have 6C3 ie 60 ways.
then for the 2 y's , we have 3 places . so selecting two places out of three places is, 3C2 ie 3 ways
this gives us 180 ways
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Permutation and combination
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regor60
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6C3 is 20, not 60, otherwise correctApoorva@5 wrote:Why cannot we solve the question by the following steps :
out of 6 places, first wefill all the X's. For that we have 6C3 ie 60 ways.
then for the 2 y's , we have 3 places . so selecting two places out of three places is, 3C2 ie 3 ways
this gives us 180 ways
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MartyMurray
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Hi Apoorva@5.Apoorva@5 wrote:Each of the [6] squares above is to have exactly one letter and nothing else placed inside it. If 3 of the letters are to be the letter X, 2 of the letters are to be the letter Y and 1 of the letters is to be the letter Z, in how many different arrangements can the squares have letters placed in them?
(A) 30
(B) 60
(C) 108
(D) 120
(E) 720
Why cannot we solve the question by the following steps :
out of 6 places, first wefill all the X's. For that we have 6C3 ie 60 ways.
then for the 2 y's , we have 3 places . so selecting two places out of three places is, 3C2 ie 3 ways
this gives us 180 ways
6C3 = 6 x 5 x 4/3! = 120/6 = 20
Having said that, I would do 6P6 then just divide by 3! and 2! to account for the repeated letters.
6!/3!2! = 60
The correct answer is B.
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Brent@GMATPrepNow
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When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:Each of the [6] squares above is to have exactly one letter and nothing else placed inside it. If 3 of the letters are to be the letter X, 2 of the letters are to be the letter Y and 1 of the letters is to be the letter Z, in how many different arrangements can the squares have letters placed in them?
(A) 30
(B) 60
(C) 108
(D) 120
(E) 720
If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]
So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]
---------------------------------------
Likewise, for this question, we can calculate the number of arrangements of the letters in XXXYYZ as follows:
There are 6 letters in total
There are 3 identical X's
There are 2 identical Y's
So, the total number of possible arrangements = 6!/[(3!)(2!)]
= 60
= B
Here are two related questions:
- https://www.beatthegmat.com/p-c-t202402.html
- https://www.beatthegmat.com/permutation- ... 81906.html
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Brent
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Hi Apoorva@5,
When dealing with Permutation question that involve 'duplicate elements', you have to be careful about how you go about your calculations. Consider this simple example:
If you have the letters A, B and C, then how many different 3-letter arrangements are there? Mathematically, you can calculate it as (3)(2)(1) = 6. You can also list out the options:
ABC
ACB
BAC
BCA
CAB
CBA
Now, what if you had A, A, and A instead. Here, the three elements are identical. There are no longer 6 possible arrangements - there's just 1...
AAA
From a math standpoint, the way to eliminate those duplicate options is to divide by a factorial of any duplicates. Since there are 3 A's, we have to divide by 3!....
(3)(2)(1)/3! = 1
GMAT assassins aren't born, they're made,
Rich
When dealing with Permutation question that involve 'duplicate elements', you have to be careful about how you go about your calculations. Consider this simple example:
If you have the letters A, B and C, then how many different 3-letter arrangements are there? Mathematically, you can calculate it as (3)(2)(1) = 6. You can also list out the options:
ABC
ACB
BAC
BCA
CAB
CBA
Now, what if you had A, A, and A instead. Here, the three elements are identical. There are no longer 6 possible arrangements - there's just 1...
AAA
From a math standpoint, the way to eliminate those duplicate options is to divide by a factorial of any duplicates. Since there are 3 A's, we have to divide by 3!....
(3)(2)(1)/3! = 1
GMAT assassins aren't born, they're made,
Rich
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Danny@GMATAcademy
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The calculation error would be harder to make if you worked from simplest item to most complex:
Ways to deal with Z: 6
Ways to deal with Ys (assuming Z already placed): 5C2 = 10
Ways to deal with Xs (assuming Z and Ys already placed): 1
So 6*10*1 = 60
Ways to deal with Z: 6
Ways to deal with Ys (assuming Z already placed): 5C2 = 10
Ways to deal with Xs (assuming Z and Ys already placed): 1
So 6*10*1 = 60
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