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10 Dec 2013, 02:09
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I have learned factorial and I can answer simple questions like how many 3-man group from 7 people.
But I become confused when it gets harder.
I must have missed the underlying fundamental concept I hope someone can shed some light.
Let's begin with very downright basic example, like:
Fill in 3 empty fields each with one of the following alphabets. How many unique combination? Order DOES matter?
Note: I don't just want the answer, I care more about how we go about tackling such question!
1) ABCDE
2) AABBCDE
3) AAABCCDDEEE
But I become confused when it gets harder.
I must have missed the underlying fundamental concept I hope someone can shed some light.
Let's begin with very downright basic example, like:
Fill in 3 empty fields each with one of the following alphabets. How many unique combination? Order DOES matter?
Note: I don't just want the answer, I care more about how we go about tackling such question!
1) ABCDE
2) AABBCDE
3) AAABCCDDEEE
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11 Dec 2013, 14:43
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ghostie
Dear ghostie
I'm happy to help.
First of all, here are some blogs that might help. This one explains the Fundamental Counting Principle, a pivotal idea.https://magoosh.com/gmat/2012/gmat-quant-how-to-count/
This one talks more about difficult counting problems.
https://magoosh.com/gmat/2013/difficult- ... -problems/
Even more so than in other areas of math, memorization is an approach that will really hamper you in counting problems. If your primary problem-solving strategy is just asking which formula to use, then you are more or less shutting yourself out of the 700 level problems. Counting problems are all about how we frame the problem, the perspective we use to interpret it. Developing this perspective is everything, and apart from that, the formulas are not very useful beyond the simple problems.
Your question doesn't quite make sense --- "Fill in 3 empty fields each with one of the following alphabets." I assume you mean
1) How many unique 5-letter combinations can we make with ABCDE?
2) How many unique 7-letter combinations can we make with AABBCDE?
3) How many unique 11-letter combinations can we make with AAABCCDDEEE?
Those are the questions I will answer.
The first, of course, is a straightforward permutation question. There are 5! = 120 unique combination.
For the other two, we will figure out the total number of permutations, and then go back and divide off repetition.
In #2, we have seven elements, so start out just by finding all permutation --- 7!. Now, if we interchange the two A's, we get the same combination, so divide by 2. Similarly, with the two B's, so divide by 2 again. There are 7!/(2*2) = 1260 unique combinations.
In #3, we have eleven elements. All permutations = 11! Now, to eliminate repetitions, we divide out for
all orders of 3 A's --- divide by 3! = 6
all orders of 2 C's --- divide by 2
all orders of 2 D's --- divide by 2
all orders of 3 E's --- divide by 3! = 6
That's 11!/(6*2*2*6) = 11!/144 = 277,200 unique combinations.
I got this last number with a calculator. It is unlikely the GMAT would expect you do the calculation in #2 --- they might leave the answer in the form 7!/4 ---- and they definitely would not expect you to come up with the answer to #3 without a calculator --- if they asked that question, it would be in the form 11!/144.
Again, don't get stuck on formulas. Perspective is everything.
Does all this make sense?
Mike
11 Dec 2013, 21:41
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ghostie
My understanding of your question is a little different. Do you mean "How many 3 letter words can you make?" Note that whether order matters depends on the exact question. If the question is "How many 3 letter words can you make?" then of course the order matters. ACB and ABC are two different words.
If instead you mean "In how many ways can you select 3 letters from the given letters?" then order doesn't matter - it is only a selection i.e. only a group of 3 letters. They are not arranged in any way.
Look at the case when order matters.
1) ABCDE
___ ___ ___
You have 5 options for the first blank, 4 options for the second blank (after you pick one for the first blank) and then only 3 options left for the third blank.
Total words possible = 5*4*3
or another way to look at it is this: select 3 letters out of 5 and arrange them. 5C3 * 3! = 5*4*3
2) AABBCDE
Here you need to make cases:
Case 1: Words with all letters unique - Select 3 of the 5 letters and arrange them (same as above) in 5C3 * 3! ways
Case 2: Words with 2 letters same - Select the letter that you want to repeat. This can be done in 2 ways (either A or B). Next select a letter out of the remaining 4. Now you have 3 letters two of which are same and 1 is different. You can arrange them in 3!/2! ways (we divide by 2! because 2 letters are the same)
Case 2 Total = 2* 4C1 * 3!/2!
Now you add the number of words you obtain in the two cases.
3) AAABCCDDEEE
Here there will be some more cases:
Case 1: All 3 letters are the same. This can be done in 2 ways (AAA or EEE)
Case 2: 2 letters same. Select the letter which will be same in 4 ways (A or C or D or E). Select the third letter in 4C1 ways (after one is already selected)
Now arrange these 3 letters in 3!/2! ways
Case 2 total = 4 * 4C1 * 3!/2!
Case 3: All 3 letters unique.
Select 3 of the 5 letters and arrange them in 5C3 * 3! ways
