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Manager  Status: mba here i come!
Joined: 07 Aug 2011
Posts: 182
Re: How many five-digit numbers are there, if the two leftmost  [#permalink]

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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 = $$19*5^3 = 2375$$
Math Expert V
Joined: 02 Sep 2009
Posts: 59674
How many five-digit numbers are there, if the two leftmost  [#permalink]

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

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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Current Student B
Joined: 23 May 2013
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Concentration: Technology, Healthcare
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Re: How many five-digit numbers are there, if the two leftmost  [#permalink]

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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.
Director  S
Joined: 17 Dec 2012
Posts: 623
Location: India
Re: How many five-digit numbers are there, if the two leftmost  [#permalink]

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

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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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Manager  B
Joined: 10 Mar 2013
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GMAT 1: 620 Q44 V31 GMAT 2: 690 Q47 V37 GMAT 3: 610 Q47 V28 GMAT 4: 700 Q50 V34 GMAT 5: 700 Q49 V36 GMAT 6: 690 Q48 V35 GMAT 7: 750 Q49 V42 GMAT 8: 730 Q50 V39 GPA: 3
Re: How many five-digit numbers are there, if the two leftmost  [#permalink]

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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
Intern  Joined: 19 Nov 2015
Posts: 6
Re: How many five-digit numbers are there, if the two leftmost  [#permalink]

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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
4*4*5*5*5+3*4*5*5*5+4*5*4*5*4=5100
Correct me where I went wrong
Intern  B
Joined: 28 Apr 2016
Posts: 15
Re: How many five-digit numbers are there, if the two leftmost  [#permalink]

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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.
Non-Human User Joined: 09 Sep 2013
Posts: 13737
Re: How many five-digit numbers are there, if the two leftmost  [#permalink]

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_________________ Re: How many five-digit numbers are there, if the two leftmost   [#permalink] 02 Oct 2018, 12:04

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