Uniquely designating stocks with codes -Handling duplicates?

This topic has expert replies
Legendary Member
Posts: 641
Joined: Tue Feb 14, 2012 3:52 pm
Thanked: 11 times
Followed by:8 members
Hello,

Can you please explain the following:

A certain stock exchange designates each stock with the one, two or three letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order, constitute a different code, how many different stocks is it possible to uniquely designate with these codes?

a)2951
b)8125
c)15600
d)16302
e)18278

OA: e


I was thinking that if we use 2 digit codes to represent the number of stocks then we can have something like AA, BB etc. So I was wondering why we don't divide by 2! here to eliminate duplicates since A1A2 is the same as A2A1? And similarly for 3 digit codes, divide by 3!

Thanks for your help.

Best Regards,
Sri
Last edited by gmattesttaker2 on Sat Apr 05, 2014 5:38 pm, edited 2 times in total.

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Sat Apr 05, 2014 5:24 pm
A certain stock exchange designates each stock with a one-, two- or three-letter code, where each letter is selected from the 26 letters of the alphabets. If the letter maybe repeated and if the same letters used in different order constitude a different code, how many different stock is it possible to uniquely designate with these codes?

A. 2,951
B. 8,125
C. 15,600
D. 16,302
E. 18,278
1-letter codes
26 letters, so there are 26 possible codes

2-letter codes
There are 26 options for the 1st letter, and 26 options for the 2nd letter.
So, the number of 2-letter codes = (26)(26) = 26²

3-letter codes
There are 26 options for the 1st letter, 26 options for the 2nd letter, and 26 options for the 3rd letter.
So, the number of 3-letter codes = (26)(26)(26) = 26³

So, the TOTAL number of codes = 26 + 26² + 26³

IMPORTANT: Before we perform ANY calculations, we should first look at the answer choices, because we know that the GMAT test-makers are very reasonable, and they don't care whether we're able make long, tedious calculations. Instead, the test-makers will create the question (or answer choices) so that there's an alternative approach.

The alternative approach here is to recognize that:
26 has 6 as its units digit
26² has 6 as its units digit
26³ has 6 as its units digit

So, (26)+(26²)+(26³) = (26)+(___6)+(____6) = _____8

Since only E has 8 as its units digit, the answer must be E

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Sat Apr 05, 2014 5:30 pm
gmattesttaker2 wrote:
I was thinking that if we use 2 digit codes to represent the number of stocks then we can have something like AA, BB etc. So I was wondering why we don't divide by 2! here to eliminate duplicates since A1A2 is the same as A2A1? And similarly for 3 digit codes, divide by 3!
For 2-digit codes, we have two DISTINCT letter locations. We have the FIRST letter and we have the SECOND letter. So, we don't need to divide by 2!

If you're in doubt, test your theory on a smaller set of letters. Let's say there are only 2 letters in the alphabet: A and B. How many 2-letter codes can we make:
AA
AB
BA
BB
We get 4 codes.
There are 2 options for the FIRST letter and 2 options for the SECOND letter.
So, the total number of 2-letter codes = (2)(2) = 4

Cheers,
Brent


Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image

Legendary Member
Posts: 641
Joined: Tue Feb 14, 2012 3:52 pm
Thanked: 11 times
Followed by:8 members

by gmattesttaker2 » Sat Apr 05, 2014 6:12 pm
Brent@GMATPrepNow wrote:
gmattesttaker2 wrote:
I was thinking that if we use 2 digit codes to represent the number of stocks then we can have something like AA, BB etc. So I was wondering why we don't divide by 2! here to eliminate duplicates since A1A2 is the same as A2A1? And similarly for 3 digit codes, divide by 3!
For 2-digit codes, we have two DISTINCT letter locations. We have the FIRST letter and we have the SECOND letter. So, we don't need to divide by 2!

If you're in doubt, test your theory on a smaller set of letters. Let's say there are only 2 letters in the alphabet: A and B. How many 2-letter codes can we make:
AA
AB
BA
BB
We get 4 codes.
There are 2 options for the FIRST letter and 2 options for the SECOND letter.
So, the total number of 2-letter codes = (2)(2) = 4

Cheers,
Brent


Cheers,
Brent

Hello Brent,

Thank you very much for the explanations.

Best Regards,
Sri