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How many possible numbers can be built from the set 3, 5, 7,

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How many possible numbers can be built from the set 3, 5, 7, [#permalink]

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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, [#permalink]

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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
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Re: Probability Problem [#permalink]

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New post 23 Oct 2009, 15:07
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 [#permalink]

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New post 23 Oct 2009, 15:21
eresh wrote:
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?


Well technically yes. 4C3*3!=4*3!=4! and 4P3=4!.

Just noted that you said that you made this question. Really good one. Do your solution uses P instead of C? Though it wont affect the calculations but still interesting. And what did you get as an answer?

And one more thing can I ask you kindly to check the following three thread and share your thoughts?


does-z-8-basic-q-85649.html
statistics-85605.html
factors-divisors-ds-question-from-gmatclub-challenge-set-85594.html

Thank in advance.
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Re: Probability Problem [#permalink]

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New post 23 Oct 2009, 15:51
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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 :oops:

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, [#permalink]

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Re: How many possible numbers can be built from the set 3, 5, 7, [#permalink]

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New post 25 Dec 2015, 13:18
using FCP rule,

Scenario-1: 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, [#permalink]

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New post 16 Sep 2016, 00:27
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, [#permalink]

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New post 23 Sep 2016, 05:11
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, [#permalink]

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New post 09 May 2017, 17:58
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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, [#permalink]

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New post 09 May 2017, 20:33
Re: How many possible numbers can be built from the set 3, 5, 7,   [#permalink] 09 May 2017, 20:33
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