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How many possible numbers can be built from the set 3, 5, 7,
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23 Oct 2009, 09:50
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How many possible numbers can be built from the set 3, 5, 7, 8 and 9 so that the number is greater than 7000?




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How many possible numbers can be built from the set 3, 5, 7,
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23 Oct 2009, 10:39




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Re: Probability Problem
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23 Oct 2009, 16:07
Well, I kinda made this one so forgot to add the line the numbers cannot be repeated.
In the line where you are doing 3C1*4C3*3!=72, it is actually 3C1*4P3 right?



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Re: Probability Problem
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23 Oct 2009, 16:21



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Re: Probability Problem
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23 Oct 2009, 16:51
Well, I was reading the probability staffs and was discussing with a friend. He told me of something like that and then I made it, so the credit is not fully mine I also got 192 as an answer, just wanted to be sure by posting it here. Solved it as you did. 5! for all the numbers. Then for 4 digit number in the first place I can only have 3 choices (7,8 and 9), from the rest of the 4 I can choose 3 (and order is important since 539 and 935 is different) so it is a 4P3. Overall 5!+ 3*4P3=192.



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Re: How many possible numbers can be built from the set 3, 5, 7,
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25 Dec 2015, 14:18
using FCP rule,
Scenario1: 4 digit number
    1st digit 2nd digit 3rd digit 4th digit
1st digit = 3 ways (as number has to be either 7, 8, 9) 2nd digit = 4 ways (we can select either of 4 remaining numbers) 3rd digit = 3 ways (we can select either of 3 remaining numbers) 4th digit = 2 ways (we can select either of 2 remaining numbers)
Total ways = 3 x 4 x 3 x 2 = 72 ways
scenario2: 5 digit number similarly, 1 st place = 5 ways 2nd place = 4 ways 3rd place = 3 ways 4th place = 2 ways 5th place = 1 way
Total ways = 5 x 4 x 3 x 2 x 1 = 120 ways
Total = 120 + 72 = 192 ways



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How many possible numbers can be built from the set 3, 5, 7,
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16 Sep 2016, 01:27
Bunuel wrote: eresh wrote: How many possible numbers can be built from the set 3,5,7,8 and 9 so that the number is greater than 7000? I assume the condition is that the numbers cannot repeat. If so then: If we have 4 digit number: first digit must be 7, 8 or 9. If we have 5 digit number any number can be the first digit. 4 digit number: 3C1*4C3*3!=72 (3C1 to choose 1 # from 7,8 or 9; 4C3 to choose the rest 3 numbers for 4 digit number out of 4 left; 3! # of combinations of these 3 numbers) 5 digit number: 5!=120 72+120=192 Hi Bunuel, Just wondering, I assumed no repetition, similar to your assumption. While i understand your method....just wanted to know if there is wrong with this method No of 4 digit number  3C1(for 7,8,9)* 4C1(since 7/8/9 cannot be repeated, 1 number chosen, 4 left to pick one)*3C1(similarly 2 numbers are chosen already,3 left)*2C1(3 numbers are chosen already, 2 left) Hope my solution is clear. Appreciate your help.



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Re: How many possible numbers can be built from the set 3, 5, 7,
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23 Sep 2016, 06:11
How can we make sure that when it starts with 3 or 5 wont give a number higher than 7000.ty



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Re: How many possible numbers can be built from the set 3, 5, 7,
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09 May 2017, 18:58
eresh wrote: How many possible numbers can be built from the set 3, 5, 7, 8 and 9 so that the number is greater than 7000? Bunuel: Please Untag probability
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Re: How many possible numbers can be built from the set 3, 5, 7,
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09 May 2017, 21:33




Re: How many possible numbers can be built from the set 3, 5, 7, &nbs
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09 May 2017, 21:33






