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How many five-digit numbers are there, if the two leftmost

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Senior Manager
Joined: 02 Dec 2007
Posts: 422
How many five-digit numbers are there, if the two leftmost [#permalink]

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27 Aug 2008, 09:39
. How many five-digit numbers are there, if the two leftmost digits are even, the other digits are odd and the digit 4 cannot appear more than once in the number?

a) 1875
b) 2000
c) 2375
d) 2500
e) 3875

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Director
Joined: 12 Jul 2008
Posts: 511
Schools: Wharton

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27 Aug 2008, 09:42
Nihit wrote:
. How many five-digit numbers are there, if the two leftmost digits are even, the other digits are odd and the digit 4 cannot appear more than once in the number?

a) 1875
b) 2000
c) 2375
d) 2500
e) 3875

C

# of five digit numbers w/o 4 = 3*4*5*5*5 = 1500
# of five digit numbers w/ 4 as first digit = 1*4*5*5*5 = 500
# of five digit number w/ 4 as second digit = 3*1*5*5*5 = 375

Total = 1500+500+375
SVP
Joined: 07 Nov 2007
Posts: 1732
Location: New York

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27 Aug 2008, 09:46
Nihit wrote:
. How many five-digit numbers are there, if the two leftmost digits are even, the other digits are odd and the digit 4 cannot appear more than once in the number?

a) 1875
b) 2000
c) 2375
d) 2500
e) 3875

(4*5 -1)*5*5*5 =2375
_________________

Smiling wins more friends than frowning

Retired Moderator
Joined: 18 Jul 2008
Posts: 897

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27 Aug 2008, 16:11
Can you tell my why I'm wrong?

Even numbers are 0,2,4,6,8
Odd numbers are 1,3,5,7,9

# of five digit numbers w/o 4 = 3*4*5*5*5 = 1500

- w/o 4, I'd get = 4*4*5*5*5

# of five digit numbers w/ 4 as first digit = 1*4*5*5*5 = 500

For this, I get 1*4*5*5*5

# of five digit number w/ 4 as second digit = 3*1*5*5*5 = 375

For this, I get 4*1*5*5*5

zoinnk wrote:
Nihit wrote:
. How many five-digit numbers are there, if the two leftmost digits are even, the other digits are odd and the digit 4 cannot appear more than once in the number?

a) 1875
b) 2000
c) 2375
d) 2500
e) 3875

C

# of five digit numbers w/o 4 = 3*4*5*5*5 = 1500
# of five digit numbers w/ 4 as first digit = 1*4*5*5*5 = 500
# of five digit number w/ 4 as second digit = 3*1*5*5*5 = 375

Total = 1500+500+375
Retired Moderator
Joined: 18 Jul 2008
Posts: 897

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27 Aug 2008, 18:01
I didn't think about the zero. Thanks!

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Re: five digit number   [#permalink] 27 Aug 2008, 18:01
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How many five-digit numbers are there, if the two leftmost

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