Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

How many 5 digit numbers can be created if the following [#permalink]
16 Sep 2005, 16:44

How many 5 digit numbers can be created if the following terms apply: the leftmost digit is even, the second is odd, the third is a non even prime and the fourth and fifth are two random digits not used before in the number?

a) 2520
b) 3150
c) 3360
d) 6000
e) 7500 _________________

the leftmost number is even. so we have 5 choices (they didn't say anything about not using 0)
the second number is odd, again we have 5 choices
third is a non even prime number, ie 3 choices (3,5,7)
fourth is digit that is not used before so total of 7 choices
and fifth, not used before so 6 chices
therefore, 5*5*3*7*6=3150

The OA is A...richardj is correct. You should not include zero as an option for the first digit.

I fell victim to this very mistake. It is frustrating because the problem asks for all even digits for the first digit, so I included zero. I got a problem wrong a week ago for not including zero when it asked for all even digits. What gives man! _________________

I agree A is the best of the possible answers here, but I do not think it is correct. Please tell me if I am wrong:

If we proceed with 4*5*3*7*6 we are not considering the case when the second and third digit are the same.

My solution would be:

(4*5*3-4*3)*7*6 + 4*3*8*7 = 2688

-4*3 because out of the 4*5*3 possibilities, 4*3 of them have the second and third digit equal.
4*3*8*7 because when the second and third digit are equal you have 8 different numbers that have not been used before.

I agree A is the best of the possible answers here, but I do not think it is correct. Please tell me if I am wrong:

If we proceed with 4*5*3*7*6 we are not considering the case when the second and third digit are the same.

My solution would be:

(4*5*3-4*3)*7*6 + 4*3*8*7 = 2688

-4*3 because out of the 4*5*3 possibilities, 4*3 of them have the second and third digit equal. 4*3*8*7 because when the second and third digit are equal you have 8 different numbers that have not been used before.

Why do we need to remove numbers where 2nd 3rd digits are same? The question didnt ask it..did it?

2nd digit can be one of 5 (1, 3, 5, 7, 9)
3rd didit can be one among (3, 5, 7)

there can be 5 * 3 = 15 arrangements between them. and all of these are DIFFERENT. So we need to include ALL of them.

I agree A is the best of the possible answers here, but I do not think it is correct. Please tell me if I am wrong:

If we proceed with 4*5*3*7*6 we are not considering the case when the second and third digit are the same.

My solution would be:

(4*5*3-4*3)*7*6 + 4*3*8*7 = 2688

-4*3 because out of the 4*5*3 possibilities, 4*3 of them have the second and third digit equal. 4*3*8*7 because when the second and third digit are equal you have 8 different numbers that have not been used before.

Why do we need to remove numbers where 2nd 3rd digits are same? The question didnt ask it..did it?

2nd digit can be one of 5 (1, 3, 5, 7, 9) 3rd didit can be one among (3, 5, 7)

there can be 5 * 3 = 15 arrangements between them. and all of these are DIFFERENT. So we need to include ALL of them.

Duttsit,
If second and third digits are the same then, only two numbers are used up till we reach fourth digit, so there is more possibility for 4 and 5 th digits that is why it has to be considered seperately.

I agree with jdotmotio...

60 possibilty for first three digit combo, out of that 12 will have second and third digit repeated so

12*8*7 + 48*7*6 = 2688 should be the answer to tis question or the question should say "no digit can be repeated"

Duttsit, If second and third digits are the same then, only two numbers are used up till we reach fourth digit, so there is more possibility for 4 and 5 th digits that is why it has to be considered seperately.

I agree with jdotmotio...

60 possibilty for first three digit combo, out of that 12 will have second and third digit repeated so

12*8*7 + 48*7*6 = 2688 should be the answer to tis question or the question should say "no digit can be repeated"

the leftmost number is even. so we have 5 choices (they didn't say anything about not using 0) the second number is odd, again we have 5 choices third is a non even prime number, ie 3 choices (3,5,7) fourth is digit that is not used before so total of 7 choices and fifth, not used before so 6 chices therefore, 5*5*3*7*6=3150

I really need someone to help me on thins:
This is how I calculated :
- 0 cannt be there in left most place. The no will become 4 digit in that case.
- Case 1:
Second and Third digit have diff number:
4C1*5C1*2C1 (7C1*6C1)

- case 2:
Both are same digits :
4C1*5C1*1 (8C1*7C1)
This is becoz only 2 digits have been used already.

Re: How many 5 digit numbers can be created if the following [#permalink]
20 Sep 2012, 00:40

The first place may be filled by 2,4,6,8 = 4 Second by 1,3,5,7,9 = 5 Third by 3,5,7 = 3 So, the total no. of combns for the three places are 4*5*3 = 60. Given there are no restrictions on balance two places, they may be filled by 10 digits each. Hence the maximum no. of combinations turns out to be 60*10*10 = 6000. However, there are restrictions, hence options d & e are ruled out. Also, the answer has to be a multiple of 60, it rules out the option b as well. Again, provided all digits of the number are different, fourth place can be filled by 7 and the fifth by 6 digits. So, in this case the number of numbers will be 60*7*6 = 60*42 = 2520. However, the restriction on digits not being repeated is only on fourth and fifth place, the number of numbers created will be more than this. So, by this methodology, I arrive at the answer of 3360 numbers. Tell me if I am wrong.

gmatclubot

Re: How many 5 digit numbers can be created if the following
[#permalink]
20 Sep 2012, 00:40

I´ve done an interview at Accepted.com quite a while ago and if any of you are interested, here is the link . I´m through my preparation of my second...

It’s here. Internship season. The key is on searching and applying for the jobs that you feel confident working on, not doing something out of pressure. Rotman has...