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

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How many five-digit numbers are there, if the two leftmost [#permalink]  03 Dec 2007, 02:51
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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
[Reveal] Spoiler: OA
CEO
Joined: 17 Nov 2007
Posts: 3577
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
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Kudos [?]: 2168 [2] , given: 359

2
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Expert's post
C.

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

where
4 cases of first digit {2,4,6,8}
5 cases of second digit {0,2,4,6,8}
1 case of 44 for two leftmost digit
5 cases of third digit {1,3,5,7,9}
5 cases of fourth digit {1,3,5,7,9}
5 cases of fifth digit {1,3,5,7,9}
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walker wrote:
C.

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

where
4 cases of first digit {2,4,6,8}
5 cases of second digit {0,2,4,6,8}
1 case of 44 for two leftmost digit
5 cases of third digit {1,3,5,7,9}
5 cases of fourth digit {1,3,5,7,9}
5 cases of fifth digit {1,3,5,7,9}

Very nice. I was stupid enough to make the mistake of forgetting 44 (4*4*5*5*5 = 2000).

Walker you're an ace in quant buddy - how the hell do you study for it? It'll be great if you start a topic 'My Quant Prep Strategies' on the 'GMAT Strategies' section
CEO
Joined: 17 Nov 2007
Posts: 3577
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
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Followers: 408

Kudos [?]: 2168 [0], given: 359

Expert's post
GK_Gmat wrote:
Walker you're an ace in quant buddy - how the hell do you study for it? It'll be great if you start a topic 'My Quant Prep Strategies' on the 'GMAT Strategies' section

Thanks, GK_Gmat. you are all here very nice in GMAT .
95% of my preparation - studying in past. I finished physicals mathematical school, have Master of Science in Physics, Master of Science in Finance, and PhD in nannotech.... But I'll try to help anyone on this forum and I hope you will help me in Verbal
Senior Manager
Joined: 09 Oct 2007
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walker wrote:
95% of my preparation - studying in past. I finished physicals mathematical school, have Master of Science in Physics, Master of Science in Finance, and PhD in nannotech.... But I'll try to help anyone on this forum and I hope you will help me in Verbal

Ok, this explains it. Now I don't feel as stupid, just chose different path
CEO
Joined: 21 Jan 2007
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Kudos [?]: 354 [0], given: 4

very curious why you are pursuing an MBA after you have a PhD and dual MS degrees? Are you aiming for science venture capital?
CEO
Joined: 17 Nov 2007
Posts: 3577
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
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Kudos [?]: 2168 [0], given: 359

Expert's post
bmwhype2 wrote:
very curious why you are pursuing an MBA after you have a PhD and dual MS degrees? Are you aiming for science venture capital?

Oh...It's a long story
SVP
Joined: 29 Aug 2007
Posts: 2496
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Kudos [?]: 562 [0], given: 19

= (4x5x5x5x5) - (1x1x5x5x5)
= 2,375

same as walker's.

walker wrote:
C.

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

where
4 cases of first digit {2,4,6,8}
5 cases of second digit {0,2,4,6,8}
1 case of 44 for two leftmost digit
5 cases of third digit {1,3,5,7,9}
5 cases of fourth digit {1,3,5,7,9}
5 cases of fifth digit {1,3,5,7,9}
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Joined: 07 Nov 2007
Posts: 1828
Location: New York
Followers: 29

Kudos [?]: 551 [0], given: 5

Re: PS - Combination: How many digit numbers??? [#permalink]  25 Aug 2008, 13:00
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alohagirl 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

EEOOO

= (combinations excludes 4 in first poistion) + (combinations with first position 4)
$$= (3*5)*(5*5*5)+(1*4)*5*5*5$$
$$= 19*(5^3)= 2375$$
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Re: [#permalink]  25 Aug 2008, 23:25
walker wrote:
C.

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

where
4 cases of first digit {2,4,6,8}
5 cases of second digit {0,2,4,6,8}
1 case of 44 for two leftmost digit
5 cases of third digit {1,3,5,7,9}
5 cases of fourth digit {1,3,5,7,9}
5 cases of fifth digit {1,3,5,7,9}

why did u do 4*5-1
and not 5*4 (as we can choose 5 digits for the first one and four for the second, or 4 for teh first one, and five for second, and either way we get 20 combinations?)

on thinkgin, is ti coz u cannot have a zero as the first digit?
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Re: PS - Combination: How many digit numbers??? [#permalink]  25 Aug 2008, 23:35
alohagirl 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

Number of combo with 4 as first digit + Number of combos with 4 as second digit = 1*4*5*5*5 + 3*5*5*5 = 2375
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Re: PS - Combination: How many digit numbers??? [#permalink]  25 Aug 2008, 23:46
I need some help with this. Why isn't it 3*5*5*5*5 + 4*4*5*5*5 ?

That is, number of cases where 4 is allowed to appear as the second digit plus number of cases where 4 is allowed to appear as the first digit.
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Re: Re: [#permalink]  26 Aug 2008, 05:23
sset009 wrote:
walker wrote:
C.

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

where
4 cases of first digit {2,4,6,8}
5 cases of second digit {0,2,4,6,8}
1 case of 44 for two leftmost digit
5 cases of third digit {1,3,5,7,9}
5 cases of fourth digit {1,3,5,7,9}
5 cases of fifth digit {1,3,5,7,9}

why did u do 4*5-1
and not 5*4 (as we can choose 5 digits for the first one and four for the second, or 4 for teh first one, and five for second, and either way we get 20 combinations?)

on thinkgin, is ti coz u cannot have a zero as the first digit?

No. that is because 4 cannot appear twice in the number.. so we are subtracting 4 4 combination.

4*5 --- because we cann't have zero's in the first digit
otherwise it would have been 5*5
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Re: PS - Combination: How many digit numbers??? [#permalink]  26 Aug 2008, 08:06
Ans 2375.

1st case: 1st digit is either 2,6,8, 2nd digit is either 2,4,6,8,0. 3rd, 4th, 5ht can be either 1,3,5,7,9.

3.5.5.5.5=1875

2nd case: 1st digit is 4, 2nd digit is 0,2,6,8. the rest are all odd digits
1.4.5.5.5=500
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Re: [#permalink]  08 Aug 2009, 09:00
walker wrote:
GK_Gmat wrote:
Walker you're an ace in quant buddy - how the hell do you study for it? It'll be great if you start a topic 'My Quant Prep Strategies' on the 'GMAT Strategies' section

Thanks, GK_Gmat. you are all here very nice in GMAT .
95% of my preparation - studying in past. I finished physicals mathematical school, have Master of Science in Physics, Master of Science in Finance, and PhD in nannotech.... But I'll try to help anyone on this forum and I hope you will help me in Verbal

Walker, I think I read your blog/posting about Chicago Booth interiew somewhere. Had the feeling that you didn't like Kellog much ( or You are choosing Chigaco since there are many Ph.D's and Novel prize winner) Anyway, I have great respect for Russian ppl...you rock lady!!!
CEO
Joined: 17 Nov 2007
Posts: 3577
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Schools: Chicago (Booth) - Class of 2011
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Kudos [?]: 2168 [0], given: 359

Re: Re: [#permalink]  08 Aug 2009, 23:49
Expert's post
Walker, I think I read your blog/posting about Chicago Booth interiew somewhere. Had the feeling that you didn't like Kellog much ( or You are choosing Chigaco since there are many Ph.D's and Novel prize winner) Anyway, I have great respect for Russian ppl...you rock lady!!!

I didn't consider Kellogg as I didn't feel the school is so famous here as Chicago. It's rather an intuitive argument than a reason.
By the way, I'm Ukrainian
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Re: Re: [#permalink]  09 Aug 2009, 19:10
@walker,

I wish to visit Ukrain someday....
btw, i am little older ( did i say little older..i am 36+), and you know what I feel like ppl will be laughing at me if i say i want to do MBA. Even so true if i say i want it from top 15 school...i dont know how the schools and interviewer will see it..i am not an overacheiver either..i am a medicore guy with a dream... sometimes i laugh at myself when i think pragmatically...
I know you have been researching/researched many school'sif I have to seek your advice/feedback ..what you will you say?

my b/g: undergrad engineering, MS computer sc..IT 5+ yrs exp, total exp 8 yrs...ya, I was in between jobs for few yrs, as it is very hard to find job in my home country unless you can pay lot of bribe or you have good connection.

walker wrote:
GK_Gmat wrote:
Walker you're an ace in quant buddy - how the hell do you study for it? It'll be great if you start a topic 'My Quant Prep Strategies' on the 'GMAT Strategies' section

Thanks, GK_Gmat. you are all here very nice in GMAT .
95% of my preparation - studying in past. I finished physicals mathematical school, have Master of Science in Physics, Master of Science in Finance, and PhD in nannotech.... But I'll try to help anyone on this forum and I hope you will help me in Verbal

Walker, I think I read your blog/posting about Chicago Booth interiew somewhere. Had the feeling that you didn't like Kellog much ( or You are choosing Chigaco since there are many Ph.D's and Novel prize winner) Anyway, I have great respect for Russian ppl...you rock lady!!!
CEO
Joined: 17 Nov 2007
Posts: 3577
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
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Kudos [?]: 2168 [0], given: 359

Re: Re: [#permalink]  09 Aug 2009, 21:29
Expert's post
I wish to visit Ukrain someday....

Just let me know and If I'm in Ukraine, we definitely will meet.

btw, i am little older ( did i say little older..i am 36+), and you know what I feel like ppl will be laughing at me if i say i want to do MBA. Even so true if i say i want it from top 15 school...i dont know how the schools and interviewer will see it..i am not an overacheiver either..i am a medicore guy with a dream... sometimes i laugh at myself when i think pragmatically...
I know you have been researching/researched many school'sif I have to seek your advice/feedback ..what you will you say?

I'm 34. My research was simple: 10 top schools minus a few schools and that's all. I strongly believe that all will more depend on me, not on the school.

Answer the main question: Why do you need MBA? Answer it honestly to yourself. And ask another question: Why not EMBA?
In order to compete with young guys, you should have "Wow" accomplishments. Let's say to be CEO of IBM or the winner of the Olympic Games. Additionally, don't overestimate GMAT. Yeah, you need 750+ but without "wow" accomplishments it won't help you too much.

Anyway, good luck in your non-easy way!
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Re: PS - Combination: How many digit numbers??? [#permalink]  27 Sep 2009, 20:49
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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

Soln: we need to break this up into multiple parts and then solve.

first to deal with digit places 3,4,5, They can be filled with odd integers (1,3,5,7,9) in 5 * 5 * 5 ways.

Now considering the left most two digits.
1) Assuming 4 has been put in the first digit place, any of the remaining 4 even numbers (2,6,8,0) can go in the second place.
Hence we have = 4 * 5 * 5 * 5 ways
2) Now if 4 goes into second digit place, any of the remaining 3 even numbers (2,6,8) can go in the first place. We cannot put 0 because we need 5 digit numbers.
Hence we have = 3 * 5 * 5 * 5 ways
3) Now if 4 is not chosen for any of the digits, we can fill the first digit place with any of the 3 even numbers (2,6,8) and the second digit place with any of the 4 even numbers (2,6,8,0)
Hence we have = 12 * 5 * 5 * 5 ways

Total number of ways
= 4 * 5 * 5 * 5 + 3 * 5 * 5 * 5 + 12 * 5 * 5 * 5
= 19 * 125
= 2375
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Re: PS - Combination: How many digit numbers??? [#permalink]  27 Sep 2009, 21:02
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

Soln: we need to break this up into multiple parts and then solve.

first to deal with digit places 3,4,5, They can be filled with odd integers (1,3,5,7,9) in 5 * 5 * 5 ways.

Now considering the left most two digits.
1) Assuming 4 has been put in the first digit place, any of the remaining 4 even numbers (2,6,8,0) can go in the second place.
Hence we have = 4 * 5 * 5 * 5 ways
2) Now if 4 goes into second digit place, any of the remaining 3 even numbers (2,6,8) can go in the first place. We cannot put 0 because we need 5 digit numbers.
Hence we have = 3 * 5 * 5 * 5 ways
3) Now if 4 is not chosen for any of the digits, we can fill the first digit place with any of the 3 even numbers (2,6,8) and the second digit place with any of the 4 even numbers (2,6,8,0)
Hence we have = 12 * 5 * 5 * 5 ways

Total number of ways
= 4 * 5 * 5 * 5 + 3 * 5 * 5 * 5 + 12 * 5 * 5 * 5
= 19 * 125
= 2375

I think there is no much need to break down in multiple parts. The solution is straight: $$[4C1*5C1-1]*(5C1)^3$$
Re: PS - Combination: How many digit numbers???   [#permalink] 27 Sep 2009, 21:02

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