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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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A. 5,760
B. 7,290
C. 7,680
D. 8,640
E. 10,240
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ScottTargetTestPrep
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Bunuel
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.
Answer: E
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Bunuel
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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General Discussion
Bunuel
Check Constructing Numbers, Codes and Passwords in our Special Questions Directory.
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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
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
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Bunuel
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
Posts: 11
VeritasKarishma
Could somebody please explain why arrange the Ds and Es is 6!/(3!*3!)?
VeritasKarishma
VeritasKarishma
Can you pls explain me one thing here, when we talk about DDD it is possible that it is 557 having 2 digits same or 777 having 3 digits same and similar is the case for EEE, so how does 6!/3!3! (which considers DDD is say 555 and EEE is say 111) account for when only 2 digits are same
Thanks in advance!
GDT
DDD is just a three digit number, whatever the digits may be (557 or 577 or 777 etc). We have arranged the 3 digits in all acceptable ways. Now we do NOT have to arrange these three digits. That is why we divide by 3!
Alternatively, think of it this way: Select 3 of the 6 spots in 6C3 ways.
In these 3 spots, put 5 or 7 in 2*2*2 ways.
In rest of the three spots, put 1 or 4 or 6 or 8 in 4*4*4 = 64
Total arrangements = 6C3 * 8 * 64 = 10,240 ways
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Kinshook
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Products:
Asked: 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 ?
The number of 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 = 6C3*2^3*4^3 = 20*8*64 = 10240
IMO E
The number of 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 = 6C3*2^3*4^3 = 20*8*64 = 10240
IMO E
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