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Re: How many fivedigit numbers are there, if the two leftmost
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28 Feb 2012, 14:09
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 fivedigit numbers are there, if the two leftmost
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28 Feb 2012, 14:47
srivas wrote: How many fivedigit 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 fivedigit 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 fivedigit numbers with two 4's as leftmost digits and with three odd last digits: 44OOO  \(5*5*5=5^3\); So, the answer is \(4*5^45^3=5^3(4*51)=125*19=2375\). Answer: C. Similar question to practice: howmany5digitnosarethereifthe2leftmostdigitsare126966.htmlHope it helps.
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Re: How many fivedigit numbers are there, if the two leftmost
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12 Feb 2014, 11:27
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 fivedigit numbers are there, if the two leftmost
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12 Mar 2014, 00:02
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 fivedigit numbers are there, if the two leftmost
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15 Mar 2014, 19:25
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 fivedigit numbers are there, if the two leftmost
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10 Jul 2015, 06:24
Argh! I forgot that 0 couldn't be the first digit; read carefully, folks! 4*5^31*1*5^3 = 5^3(201)=125*19



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Re: How many fivedigit numbers are there, if the two leftmost
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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



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Re: How many fivedigit numbers are there, if the two leftmost
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02 Jun 2017, 19:13
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 = 2500125 = 2375.



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Re: How many fivedigit numbers are there, if the two leftmost
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02 Oct 2018, 12:04
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Re: How many fivedigit numbers are there, if the two leftmost
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