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[PS][Rank] HSPA posts

by HSPA » Thu Mar 31, 2011 9:29 am
With out repetition, how many possible numbers can there be >56000 with digits 4,5,6,7,8
a) 84
b) 120
C) 90
d) 360
e) None
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by srcc25anu » Thu Mar 31, 2011 9:47 am
IMO it should be C (90)
there can be 2 cases:
1) when we have 5 in first place: there can be 1*3*3*2*1 = 18 numbers
2) when we have 6,7 or 8 in first place: then there can be 3*4*3*2*1 = 72 numbers
total = 90 numbers
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by manpsingh87 » Thu Mar 31, 2011 9:53 am
HSPA wrote:With out repetition, how many possible numbers can there be >56000 with digits 4,5,6,7,8
a) 84
b) 120
C) 90
d) 360
e) None
here 2 cases are possible:

1) when 5 is at the left most position
2) when 5 is not at the left most position.

case 1) 5----,thousand's place can be filled in three ways (by any of 6,7,8), similarly hundred's place can be filled in again 3 ways, ten's place in 2 and unit in 1 way,

hence total no. of ways = 1*3*3*2*1=18

case 2)when 5 is not at the left most position, then it can be filled by any of 6,7,8 in 3 ways, and the remaining 4 places can be filled by the 4 no. in 4! ways, total no. of ways= 3*4!=72

hence total no. of ways=18+72=90
hence C
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by MAAJ » Thu Mar 31, 2011 9:56 am
Nice one srcc25anu! I was getting 72 all the time, 'cause I was missing the first part :p
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by force5 » Thu Mar 31, 2011 1:47 pm
yes no repetition gets you 90 numbers
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by HSPA » Thu Mar 31, 2011 10:26 pm
well done.. OA is pointing 90
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by GMATGuruNY » Fri Apr 01, 2011 4:32 am
HSPA wrote:With out repetition, how many possible numbers can there be >56000 with digits 4,5,6,7,8
a) 84
b) 120
C) 90
d) 360
e) None
Do only as much math as is necessary.

The first digit can be 5,6,7,8 = 4 choices.
Number of ways to arrange the remaining 4 digits = 4! = 24.
Multiplying our choices for the first digit with our choices for the remaining digits, we get 4*24 = 96 possible numbers.
A few of the numbers that begin with 5 will be less then 56,000, so we need an answer choice that is just a bit less than 96.

The correct answer is C.
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