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HSPA: Legendary Member; Posts: 1101; Joined: Fri Jan 28, 2011 7:26 amFollowed by: 13 members; Thanked: 47 times; GMAT Score:640

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

First take: 640 (50M, 27V) - RC needs 300% improvement
Second take: coming soon..
Regards,
HSPA.

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Source: Beat The GMAT — Problem Solving | Thu Mar 31, 2011 9:29 am

srcc25anu: Master | Next Rank: 500 Posts; Posts: 423; Joined: Fri Jun 11, 2010 7:59 amFollowed by: 2 members; Location: Seattle, WA; Thanked: 86 times

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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manpsingh87: Master | Next Rank: 500 Posts; Posts: 436; Joined: Tue Feb 08, 2011 3:07 amFollowed by: 6 members; Thanked: 72 times

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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MAAJ: Master | Next Rank: 500 Posts; Posts: 243; Joined: Sun Jul 12, 2009 7:12 amFollowed by: 2 members; Location: Dominican Republic; Thanked: 31 times; GMAT Score:480

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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force5: Legendary Member; Posts: 582; Joined: Tue Mar 08, 2011 12:48 amFollowed by: 6 members; Thanked: 61 times; GMAT Score:740

by force5 » Thu Mar 31, 2011 1:47 pm

yes no repetition gets you 90 numbers

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HSPA: Legendary Member; Posts: 1101; Joined: Fri Jan 28, 2011 7:26 amFollowed by: 13 members; Thanked: 47 times; GMAT Score:640

by HSPA » Thu Mar 31, 2011 10:26 pm

well done.. OA is pointing 90

First take: 640 (50M, 27V) - RC needs 300% improvement
Second take: coming soon..
Regards,
HSPA.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

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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