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Re: How many five-digit numbers are there, if the two leftmost [#permalink]

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28 Feb 2012, 14:09

1

This post received KUDOS

there are three possible patters.

1. when 4 is the 1st digit, and not the 2nd digit \(= 1*4*5*5*5 = 4*5^3\) 2. when 4 is the 2nd digit, and not the 1st digit \(= 3*1*5*5*5 = 3*5^3\) ... 1st digit can't be 0 either 1. when neither 1st nor 2nd digit is 4\(= 3*4*5*5*5 = 12*5^3\)

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

Easier approach:

Total five-digit numbers with two even leftmost digits and with three odd last digits: EEOOO - \(4*5*5*5*5=4*5^4\) (notice that we have only 4 choices for the first digit since we cannot use zero);

Total five-digit numbers with two 4's as leftmost digits and with three odd last digits: 44OOO - \(5*5*5=5^3\);

So, the answer is \(4*5^4-5^3=5^3(4*5-1)=125*19=2375\).

Re: How many five-digit numbers are there, if the two leftmost [#permalink]

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23 Sep 2013, 04:52

Hello from the GMAT Club BumpBot!

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First constraint=2 leftmost digits even. Therefore 4*5 ways of forming the 2 leftmost digits as the number cannot start with 0. Constraint within this constraint: 4 should not occur twice. We see that 4 can occur twice only once . Therefore tre are 4*5 - 1 ways of forming the 2 leftmost digits=19 ways Third constraint: The other 3 digits are odd. Therefore 5*5*5 ways of forming the 3 rightmost digits=125 ways

Total number of possible combinations given the constraints is 19*125= 2375
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Re: How many five-digit numbers are there, if the two leftmost [#permalink]

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02 May 2015, 05:58

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: How many five-digit numbers are there, if the two leftmost [#permalink]

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17 Feb 2016, 06:35

According to question Case:1 When 4th digit is even 4*4*5*5*5(when 0 is 4th digit) And when any even number other than 0 then. 3*4*5*5*5 Case:2 When 4th digit is odd 4*5*4*5*4 Adding all cases 4*4*5*5*5+3*4*5*5*5+4*5*4*5*4=5100 Correct me where I went wrong

gmatclubot

Re: How many five-digit numbers are there, if the two leftmost
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17 Feb 2016, 06:35

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