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Math Expert V
Joined: 02 Sep 2009
Posts: 55664
How many different 6-digit positive integers are there, where 3 of the  [#permalink]

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Difficulty:   65% (hard)

Question Stats: 44% (02:51) correct 56% (02:45) wrong based on 16 sessions

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How many different 6-digit positive integers are there, where 3 of the digits are each one of the digits 5 or 7, and the other 3 digits are each of one of the digits 1, 4, 6, or 8 ?

A. 5,760

B. 7,290

C. 7,680

D. 8,640

E. 10,240

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Math Expert V
Joined: 02 Sep 2009
Posts: 55664
Re: How many different 6-digit positive integers are there, where 3 of the  [#permalink]

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Bunuel wrote:
How many different 6-digit positive integers are there, where 3 of the digits are each one of the digits 5 or 7, and the other 3 digits are each of one of the digits 1, 4, 6, or 8 ?

A. 5,760

B. 7,290

C. 7,680

D. 8,640

E. 10,240

Check Constructing Numbers, Codes and Passwords in our Special Questions Directory.
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Senior Manager  P
Joined: 19 Oct 2018
Posts: 480
Location: India
Re: How many different 6-digit positive integers are there, where 3 of the  [#permalink]

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1. When there are 2 same digits and 1 different are selected from 5 and 7, and 3 different digits are selected from 1,4,6 and 8
Total numbers possible= 2*4*6!/2!=2880
2. When there are 2 same digits and 1 different selected from 5 and 7, and 2 same and 1 different are selected from 1,4,6 and 8
Total numbers possible= 2*6*2*6!/2!2!=4320
3. When there are 2 same digits and 1 different selected from 5 and 7, and 3 same digits selected from 1,4,6 and 8
Total numbers possible= 2*4*6!/2!3!=480
4. When there are 3 same digits from 5 and 7, and 3 different digits selected from 1,4,6 and 8
Total numbers possible= 2*4*6!/3!=960
5. When there are 3 same digits selected from 5 and 7, and 2 same and 1 different are selected from 1,4,6 and 8
Total numbers possible= 2*6*2*6!/3!2!=1440
6. When there are 3 same digits selected from 5 and 7, and 3 same digits selected from 1,4,6 and 8
Total numbers possible= 2*4*6!/3!3!=160

different 6-digit positive integers possible under given constraints= 2880+4320+480+960+1440+160=10240
CEO  P
Joined: 18 Aug 2017
Posts: 3868
Location: India
Concentration: Sustainability, Marketing
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WE: Marketing (Energy and Utilities)
Re: How many different 6-digit positive integers are there, where 3 of the  [#permalink]

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Bunuel wrote:
How many different 6-digit positive integers are there, where 3 of the digits are each one of the digits 5 or 7, and the other 3 digits are each of one of the digits 1, 4, 6, or 8 ?

A. 5,760

B. 7,290

C. 7,680

D. 8,640

E. 10,240

GMATinsight ; sir for this particular question , i tired solving the same but i am ended up getting many possible cases since the placement of 5 & 7 is not fixed ...
is there any easy way to solve this question <120 sec _________________
If you liked my solution then please give Kudos. Kudos encourage active discussions.
Target Test Prep Representative D
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Joined: 14 Oct 2015
Posts: 6563
Location: United States (CA)
Re: How many different 6-digit positive integers are there, where 3 of the  [#permalink]

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Bunuel wrote:
How many different 6-digit positive integers are there, where 3 of the digits are each one of the digits 5 or 7, and the other 3 digits are each of one of the digits 1, 4, 6, or 8 ?

A. 5,760

B. 7,290

C. 7,680

D. 8,640

E. 10,240

The number of ways to create the 6-digit number, if the first 3 digits are either 5 or 7 and the last 3 are 1, 4, 6, or 8 is 2 x 2 x 2 x 4 x 4 x 4 = 8 x 64 = 512. However, since there are 6!/(3!3!) = 720/36 = 20 ways to arrange the digits, there are a total 512 x 20 = 10,240 such numbers.

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Veritas Prep GMAT Instructor D
Joined: 16 Oct 2010
Posts: 9328
Location: Pune, India
Re: How many different 6-digit positive integers are there, where 3 of the  [#permalink]

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1
Bunuel wrote:
How many different 6-digit positive integers are there, where 3 of the digits are each one of the digits 5 or 7, and the other 3 digits are each of one of the digits 1, 4, 6, or 8 ?

A. 5,760

B. 7,290

C. 7,680

D. 8,640

E. 10,240

3 digits, say DDD, need to be selected from 5 and 7. You can do this in 2*2*2 ways. You get all combinations eg 555, 557, 577, 575 etc
3 digits, say EEE, need to be selected from 1, 4, 6 and 8. You can do this in 4*4*4 ways. You get all combinations e.g. 111, 141... etc
(A digit can be repeated)

Now you just need to mix up these two DDD and EEE to get a 6 digit number. You do need to arrange the Ds and Es among themselves since you have already accounted for their arrangements.

Total number of ways = 2*2*2*4*4*4 * 6!/3!*3! = 8*64*20 = 10,240

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Karishma
Veritas Prep GMAT Instructor Re: How many different 6-digit positive integers are there, where 3 of the   [#permalink] 19 May 2019, 21:16
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