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How many fourdigit numbers one can arrange using all the [#permalink]
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28 Jul 2003, 12:16
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How many fourdigit numbers one can arrange using all the digits, such that each number consists of two pairs of different digits. For example, 3377, 4545, or 1221.



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Kind boy, Stolyar, does this go in a pattern like ETS questions?
Congrats on 49 Quant?
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First digit can be any of 1 thru 9 = 9 ways
Second digit should be same as 1st digit = 1 way
Third digit can be any one of 0 thru 9 (except the digit used for 1st posn ) = 9 ways
4th digit is to be same as 3rd digit = 1
So total ways : 9 x 1 x 9 x 1 = 81
but 3rd and 4th digits may change their positionas
So total numbers = 2 x 81 = 162
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Brainless wrote: First digit can be any of 1 thru 9 = 9 ways Second digit should be same as 1st digit = 1 way Third digit can be any one of 0 thru 9 (except the digit used for 1st posn ) = 9 ways 4th digit is to be same as 3rd digit = 1 So total ways : 9 x 1 x 9 x 1 = 81 but 3rd and 4th digits may change their positionas So total numbers = 2 x 81 = 162
You are close but the 2nd digit does not necessarily have to match the first one.
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Is it 270 ?
9x1x10x1 = 90
9x10x1x1 = 90
9x1x1x10 = 90
Total = 270



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kpadma wrote: Is it 270 ?
9x1x10x1 = 90 9x10x1x1 = 90 9x1x1x10 = 90
Total = 270
Not quite. Much closer. Rethink your approach carefully.
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It should be either 180 or 360? But which one?



Director
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After thinking a while, I think the answer is 180!
I need multiples of 75 minutes to finish the quant if I slove the
problems in such a speed



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510.
My approach
Let's count all possible combinations, including those that begin with 0.
There are 6 ways to arrange a number, f.e. 3377, 3737, 3773, 7733, 7373, 7337.
And there are 10*1*9*1=90 ways to combine such number from digits.
Final step is to subtract numbers, beginning with 0
1*10*1*1*3=30
The ans is 6*9030=510



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RK73 wrote: 510. My approach Let's count all possible combinations, including those that begin with 0. There are 6 ways to arrange a number, f.e. 3377, 3737, 3773, 7733, 7373, 7337. And there are 10*1*9*1=90 ways to combine such number from digits. Final step is to subtract numbers, beginning with 0 1*10*1*1*3=30 The ans is 6*9030=510 Quote: >>And there are 10*1*9*1=90 ways to combine such number from digits. Final step is to subtract numbers, beginning with 0 You are double counting a few numbers here (for example, this method counts both 9 and 5, and 5 and 9, but they are the same because you already consider all combinations of 55 and 99. Quote: Final step is to subtract numbers, beginning with 0 1*10*1*1*3=30
Your final step is correct, but you are using wrong approach and calculation.
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Right.
Then
3*9030=240.



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RK73 wrote: Right.
Then 3*9030=240.
My Solution:
Method 1:
We have 2 different digits in each number. Let's call them A and B. Let's say we don't care about what order A and B are, so there are 10 * 9 / 2 = 45 difference pairings of A and B.
Now, let figure out how many way AA and BB can be put together. We have 4 digits, so there are 4! ways to arrange the four numbers, but the As and Bs are indistiguishable from each other so we need to adjust this by 2! twice. Hence, for a specific A and B the number of ways AABB can be combined is 4!/(2!2!) = (4*3)/(2*1) = 6.
So now we have 45 * 6 = 270 ways that AABB can be combined.
However, we cannot have 0 as a first digit. Since 0 is distributed the same as any of the other nine digits, exactly 10% of the numbers will start with zero and 90% will not. Hence the answer is 270 * .9 = 243.
Method 2:
There are 9 ways to pick the first number (excluding zero). Let's call this number A. For any given A, there are 9 ways to pick the other number B. So there are 81 ways to pick A as the first number and B as the other number. IF we set A to be the first digit, then there are 3 ways that the other 3 can be arranged: AABB, ABAB, or ABBA. Hence, there are 81 * 3 = 243 ways to make a four digit number with 2 paired numbers.
COMBINATORICS IS NOT ABOUT MEMORIZING FORMULAS, IT IS ABOUT COUNTING LOGICALLY AND SYSTEMATICALLY.
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AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993



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I invented this question occasionally, basing on the previous question in the forum. So I myself do not know the right answer.
My solution:
2 pairs can be arranged in 4!/(2!*2!)=6 ways
Now, count pairs: _ _ _ _
the first position  9 ways
any of the three left  the same digit  1 way
the two left  9*1 ways
So, 9*1*9*1=81
I feel we have to multiply 6*81=486
But some doubling effect takes place, no?
Where is it?



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Quote: the first position  9 ways any of the three left  the same digit  1 way the two left  9*1 ways
You are double counting here. Say you get a 3 first, then a 5 second. It is still posible to get a 5 first and a 3 second, but they are both counted when you did the 4C2 calculation.
Even so, your method is still sloppy, since by using the 4C2 formula, you mix number that CAN be zero, with numbers that cannot be zero.
While you might back in to the correct answer, your solution in not good.
Sorry bro.
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AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993



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AkamaiBrah wrote: Brainless wrote: First digit can be any of 1 thru 9 = 9 ways Second digit should be same as 1st digit = 1 way Third digit can be any one of 0 thru 9 (except the digit used for 1st posn ) = 9 ways 4th digit is to be same as 3rd digit = 1 So total ways : 9 x 1 x 9 x 1 = 81 but 3rd and 4th digits may change their positionas So total numbers = 2 x 81 = 162 You are close but the 2nd digit does not necessarily have to match the first one.
I should have multiplied by 3 to get total numbers. Thanks anyway .
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