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

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

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New post 28 Feb 2012, 13:09
1
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\(= 3*4*5*5*5 = 12*5^3\)

total = \(19*5^3 = 2375\)
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How many five-digit numbers are there, if the two leftmost  [#permalink]

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New post 28 Feb 2012, 13:47
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srivas 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


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 \(4*5^4-5^3=5^3(4*5-1)=125*19=2375\).

Answer: C.

Similar question to practice: how-many-5-digit-nos-are-there-if-the-2-leftmost-digits-are-126966.html

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

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New post 12 Feb 2014, 10:27
The question is super easy if you just consider the possibilities

_ 0 1 1 1
2 2 3 3 3
4 4 5 5 5
6 6 7 7 7
8 8 9 9 9

Considering the first number is NOT a 4, then we have:

(3) (5) (5) (5) (5) possibilities for each digit = 3*5^4

Considering the first number IS a 4, then we have:

(1) (4) (5) (5) (5) possibilities for each digit = 4* 5^3

3*5^4 + 4*5^3 = 3*625 + 4*125 = 1875 + 500 = 2375.

Solved in under 2 minutes.
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Re: How many five-digit numbers are there, if the two leftmost  [#permalink]

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New post 11 Mar 2014, 23:02
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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New post 15 Mar 2014, 18:25
No 4's + 1st Digit 4 + 2nd Digit 4 = Total

3*4*(5^3) + 1*4*(5^3) + 3*1*(5^3)
(5^3)(12+4+3)
(125)(19)
=2375
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Re: How many five-digit numbers are there, if the two leftmost  [#permalink]

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New post 10 Jul 2015, 05:24
Argh! I forgot that 0 couldn't be the first digit; read carefully, folks! ;)
4*5^3-1*1*5^3 = 5^3(20-1)=125*19
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Re: How many five-digit numbers are there, if the two leftmost  [#permalink]

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New post 17 Feb 2016, 05:35
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
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Re: How many five-digit numbers are there, if the two leftmost  [#permalink]

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New post 02 Jun 2017, 18:13
I found the following way much easier:
Total combinations = 4*5*5*5*5 = 2500.
Total combinations with two 4s = 5*5*5 = 125. (only odds can be arranged)
Hence, total combinations with only one 4 = 2500-125 = 2375.
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Re: How many five-digit numbers are there, if the two leftmost  [#permalink]

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