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How many 7-digit numbers can be arranged from 6690690

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How many 7-digit numbers can be arranged from 6690690 [#permalink]

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17 May 2005, 13:18
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How many 7-digit numbers can be arranged from 6690690?
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17 May 2005, 23:37
christoph wrote:
7!/(3!*2!*2!)

christoph, how about 0s when they come in the first place?
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18 May 2005, 00:18
emilem
u r rite n this is the trick in that v need to takecare off

in my opinion answer is 60

solution:- 6!/(31*2!*2!) * 2

explanation:- first v will permutate for last 6 places were v can fit in any of the three digits i.e 6,9,0 and multiply it by two for first position where only two digits can come
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i believe that all the problem solving calculations in Quantitative section have simple calculations. No Long Mahcine requiring calculations r required :D

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18 May 2005, 02:30
emilem wrote:
christoph wrote:
7!/(3!*2!*2!)

christoph, how about 0s when they come in the first place?

!?** you are right ! is it 150 ? 6!/(2!*2!*2!) + 6!/(3!*2!)
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If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

Last edited by christoph on 18 May 2005, 06:58, edited 1 time in total.
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18 May 2005, 06:15
emilem wrote:
How many 7-digit numbers can be arranged from 6690690?

0=2times
6=3times
9=2times

first digit can't be zero, so that leaves with any of the five digits:
5*6!/2!*3!*2!
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18 May 2005, 06:15
do we consider a case when there are 2 "O"s at the beginning?

7!/(3! 2! 2!) - 6!/(3! 2! 1!) -5!/(3! 2!) ????
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18 May 2005, 12:56
july05 wrote:
do we consider a case when there are 2 "O"s at the beginning?

7!/(3! 2! 2!) - 6!/(3! 2! 1!) -5!/(3! 2!) ????

julie

I don't think you need to include a case with 2 zeros up front becasue those cases are already included in 6!/(3! 2! 1!)
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19 May 2005, 02:49
Folks, the OA is 150. The interesting method for solving this problem that I came across in a book is that first we need to find total no. of possibilities, which is 7!/(3!*2!*2!)=210. Then, assess the position of 0s. Since we have 2 zeros out of 7 digits that can stand in the first place, 2/7th of total possibilities will include 0 in the first place. Therefore, eliminating them we have 5/7th of total possibilities or 5/7*210=150.

Do you think, guys, this solution is applicable to all this kind of problems? Just by using fractions, it seems pretty easy and logical, but would like to see your comments. By the way, other solutions are really good.
19 May 2005, 02:49
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