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 500,000 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:

A 5-digit code consists of one number digit chosen from 1, 2 [#permalink]

Show Tags

08 May 2012, 23:30

1

This post received KUDOS

20

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

48% (03:05) correct
52% (01:55) wrong based on 507 sessions

HideShow timer Statistics

A 5-digit code consists of one number digit chosen from 1, 2, 3 and four letters chosen from A, B, C, D, E. If the first and last digit must be a letter digit and each digit can appear more than once in a code, how many different codes are possible?

Choose one number from 3 numbers in 3C1 = 3 ways Choose one position from the middle three for the number in 3C1 = 3 ways The other four positions can be filled by the 5 letters in 5^4 ways.

Therefore total number of codes possible = 3*3*(5^4) = 5,625

Choose one number from 3 numbers in 3C1 = 3 ways Choose four letters from 5 letters in 5C4 = 5 ways Choose one position from the middle three for the number in 3C1 = 3 ways The other four positions can be filled by the 4 letters in 4^4 ways.

Therefore total number of codes possible = 3*5*3*(4^4) = 45*16*16 = 11,520

For the first and last positions the letters can be chosen in 5c1 ways, but if we get say the same letter twice then we need to divide by 2! to avoid duplicates since we are looking for different codes.

For the 3 positions in between, we can choose 1 digit in 4c1 ways, and the remaining 2 digits in 5c1 and 5c1 ways. These three can rearrange themselves in 3! ways and we again divide by 2! to avoid duplicates since we can have the letters to repeat themselves. I have shown this below :-

A friend asked me this question recently and I wasn't able to get the official answer for this question. I am not sure about the source or the difficulty level. The questions is as follows :-

A 5-digit code consists of one number digit chosen from 1, 2, 3 and four letters chosen from A, B, C, D, E. If the first and last digit must be a letter digit and each digit can appear more than once in a code, how many different codes are possible?

A. 375 B. 625 C. 1,875 D. 3,750 E. 5,625

Notice that each digit can appear more than once in a code.

Since there should be 4 letters in a code (X-X-X-X) and each letter can take 5 values (A, B, C, D, E) then total # of combinations of the letters only is 5*5*5*5=5^4.

Now, we are told that the first and last digit must be a letter digit, so number digit can take any of the three slots between the letters: X-X-X-X, so 3 positions and the digit itself can take 3 values (1, 2, 3).

Choose one number from 3 numbers in 3C1 = 3 ways Choose four letters from 5 letters in 5C4 = 5 ways Choose one position from the middle three for the number in 3C1 = 3 ways The other four positions can be filled by the 4 letters in 4^4 ways.

Therefore total number of codes possible = 3*5*3*(4^4) = 45*16*16 = 11,520

For the first and last positions the letters can be chosen in 5c1 ways, but if we get say the same letter twice then we need to divide by 2! to avoid duplicates since we are looking for different codes.

For the 3 positions in between, we can choose 1 digit in 4c1 ways, and the remaining 2 digits in 5c1 and 5c1 ways. These three can rearrange themselves in 3! ways and we again divide by 2! to avoid duplicates since we can have the letters to repeat themselves. I have shown this below :-

You are allowed duplicates. Even if A appears in the first as well as the last position, it will give you a code different from what you get when you have different letters in the first and the last position. You need to arrange the letters here. If instead you needed to just select groups, then yes, you would have worried about the effect of duplicates.

You select a letter for the first position in 5C1 ways and a letter for the last position in 5C1 ways. Say you put the digit in the second position. You can select a digit for the second position in 3C1 ways. You can select the letters for the third and fourth positions in 5C1 and 5C1 ways. Hence, you get 5*3*5*5*5 codes. But here, we have put the digit in the second place. It could have been in the third or fourth place too. So you multiply the above given result by 3. Hence total number of codes = 5*3*5*5*5*3 = 5625
_________________

Choose one number from 3 numbers in 3C1 = 3 ways Choose four letters from 5 letters in 5C4 = 5 ways Choose one position from the middle three for the number in 3C1 = 3 ways The other four positions can be filled by the 4 letters in 4^4 ways.

Therefore total number of codes possible = 3*5*3*(4^4) = 45*16*16 = 11,520

For the first and last positions the letters can be chosen in 5c1 ways, but if we get say the same letter twice then we need to divide by 2! to avoid duplicates since we are looking for different codes.

For the 3 positions in between, we can choose 1 digit in 4c1 ways, and the remaining 2 digits in 5c1 and 5c1 ways. These three can rearrange themselves in 3! ways and we again divide by 2! to avoid duplicates since we can have the letters to repeat themselves. I have shown this below :-

You are allowed duplicates. Even if A appears in the first as well as the last position, it will give you a code different from what you get when you have different letters in the first and the last position. You need to arrange the letters here. If instead you needed to just select groups, then yes, you would have worried about the effect of duplicates.

You select a letter for the first position in 5C1 ways and a letter for the last position in 5C1 ways. Say you put the digit in the second position. You can select a digit for the second position in 3C1 ways. You can select the letters for the third and fourth positions in 5C1 and 5C1 ways. Hence, you get 5*3*5*5*5 codes. But here, we have put the digit in the second place. It could have been in the third or fourth place too. So you multiply the above given result by 3. Hence total number of codes = 5*3*5*5*5*3 = 5625

Karishma can you tell me wher I am wrong- 1st and last places can be filled in 5 X 5 ways, another 3 this way

5C1X5C1X3C1 and they can be arranged among 3! ways = 5C1X5C1X3C1 x3! = 5x5x3x3x2

finally 5x5x5x5x3x3x2
_________________

Like my post Send me a Kudos It is a Good manner. My Debrief: http://gmatclub.com/forum/how-to-score-750-and-750-i-moved-from-710-to-189016.html

Karishma can you tell me wher I am wrong- 1st and last places can be filled in 5 X 5 ways, another 3 this way

5C1X5C1X3C1 and they can be arranged among 3! ways = 5C1X5C1X3C1 x3! = 5x5x3x3x2

finally 5x5x5x5x3x3x2

I am not sure how you have worked this out.

5*5 is fine for first and last positions - they must be letters. But you need 5 digit code so you have another 3 positions to fill

* _ _ _ *

You must use one number digit so you can select a number digit in 3 ways and the position for that number digit in 3 ways. Also the other two positions must be letters so they can be selected in 5*5 ways.

Re: A 5-digit code consists of one number digit chosen from 1, 2 [#permalink]

Show Tags

05 Nov 2014, 07:58

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: A 5-digit code consists of one number digit chosen from 1, 2 [#permalink]

Show Tags

12 May 2015, 05:36

1

This post was BOOKMARKED

Can someone pls tell me if my approach is correct :

We need to fill these 5 blank spaces. _ _ _ _ _ We have 5 letters of which we need to pick 4. That makes it 5P4. Now 4 of these 5 letters could repeat too. So to account for this, we need to divide 5P4 by 4! Of the 3 numbers we need to pick 1. So 3P1. This makes it 5P4/4! * 3P1 * 5P4/4! * 5P4/4! * 5P4/4! or 5P4/4! * 5P4/4! * 3P1 * 5P4/4! * 5P4/4! or 5P4/4! * 5P4/4! * 5P4/4! * 3P1 * 5P4/4!

Accounting for all the scenarios we get 5P4/4! * 3P1 * 5P4/4! * 5P4/4! * 5P4/4! + 5P4/4! * 5P4/4! * 3P1 * 5P4/4! * 5P4/4! + 5P4/4! * 5P4/4! * 5P4/4! * 3P1 * 5P4/4!

Re: A 5-digit code consists of one number digit chosen from 1, 2 [#permalink]

Show Tags

18 Nov 2015, 03:59

1

This post received KUDOS

gmatser1 wrote:

Can someone explain why the first and last digits are 5^2 and not just 5 options?

since repetitions are allowed, there are 5 ways to choose the letters for the first slot, and five ways to choose for last slot

thats 5 x _ x _ x _ x 5

you still have to choose 1 digit and 2 letters between. For the letters, again, repetitions are allowed so each slot can be filled in 5 ways. For digit, it can only be filled in 3 ways. However, since the digit can be in any one of the 3 middle slots, so we multiply by 3 again:

5 x 3 x 5 x 5 x 5 or 5 x 5 x 3 x 5 x 5 or 5 x 5 x 5 x 3 x 5

Hey, guys, So, I’ve decided to run a contest in hopes of getting the word about the site out to as many applicants as possible this application season...

Whether you’re an entrepreneur, aspiring business leader, or you just think that you may want to learn more about business, the thought of getting your Masters in Business Administration...

Whether you’re an entrepreneur, aspiring business leader, or you just think that you may want to learn more about business, the thought of getting your Masters in Business Administration...