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How many different 6digit positive integers are there, where 3 of the
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16 May 2019, 00:00
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44% (02:51) correct 56% (02:45) wrong based on 16 sessions
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How many different 6digit 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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Re: How many different 6digit positive integers are there, where 3 of the
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16 May 2019, 00:02
Bunuel wrote: How many different 6digit 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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Re: How many different 6digit positive integers are there, where 3 of the
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16 May 2019, 00:48
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 6digit positive integers possible under given constraints= 2880+4320+480+960+1440+160=10240



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Re: How many different 6digit positive integers are there, where 3 of the
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16 May 2019, 10:45
Bunuel wrote: How many different 6digit 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
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Re: How many different 6digit positive integers are there, where 3 of the
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19 May 2019, 20:09
Bunuel wrote: How many different 6digit 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 6digit 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. Answer: E
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Re: How many different 6digit positive integers are there, where 3 of the
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19 May 2019, 21:16
Bunuel wrote: How many different 6digit 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 Answer (E)
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Re: How many different 6digit positive integers are there, where 3 of the
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19 May 2019, 21:16






