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

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03 Dec 2007, 03:51

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39% (02:51) correct
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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.

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}

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

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 .
Unfortunately, I guess that my quant preparation couldn't help you. .
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

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

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}

Re: PS - Combination: How many digit numbers??? [#permalink]

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25 Aug 2008, 14:00

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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) \(= (3*5)*(5*5*5)+(1*4)*5*5*5\) \(= 19*(5^3)= 2375\)
_________________

Your attitude determines your altitude Smiling wins more friends than frowning

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?

Re: PS - Combination: How many digit numbers??? [#permalink]

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26 Aug 2008, 00:35

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

Re: PS - Combination: How many digit numbers??? [#permalink]

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26 Aug 2008, 00:46

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.
_________________

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
_________________

Your attitude determines your altitude Smiling wins more friends than frowning

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 . Unfortunately, I guess that my quant preparation couldn't help you. . 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!!!

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
_________________

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?

please be practical..

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.

canadamba wrote:

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 . Unfortunately, I guess that my quant preparation couldn't help you. . 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!!!

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

canadamba wrote:

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?

please be practical..

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!
_________________

Re: PS - Combination: How many digit numbers??? [#permalink]

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27 Sep 2009, 21:49

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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

Re: PS - Combination: How many digit numbers??? [#permalink]

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27 Sep 2009, 22:02

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

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