How to solve this Q?

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sanaa.rizwan: Senior | Next Rank: 100 Posts; Posts: 69; Joined: Wed Aug 08, 2012 9:00 amFollowed by: 1 member

by sanaa.rizwan » Mon Apr 01, 2013 3:27 pm

A researcher plans to indentify each participant in a certain medical experiment with a code consisting of either a single or a pair of distinct letters written in alphabetical order, what is the least number of letters that can be used if there are 12 participants and each participant is to receive a different code.

A.4
B.5
C.6
D.7
E.8

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Source: Beat The GMAT — Problem Solving | Mon Apr 01, 2013 3:27 pm

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by GMATGuruNY » Mon Apr 01, 2013 4:16 pm

sanaa.rizwan wrote:A researcher plans to indentify each participant in a certain medical experiment with a code consisting of either a single or a pair of distinct letters written in alphabetical order, what is the least number of letters that can be used if there are 12 participants and each participant is to receive a different code.

A.4
B.5
C.6
D.7
E.8

We can plug in the answers, which represent the LEAST number of letters than can be used.

Answer choice A: 4
Let the 4 letters be A, B, C and D.
Codes that can be formed:
A
B
C
D
AB
AC
AD
BC
BD
CD
Total codes = 10.
Since 12 codes are needed, eliminate A.

Answer choice B: 5
Let the 5 letters be A, B, C, D and E.
Added to the list of 10 codes above will be the following codes:
E
AE
We can stop here, since the total number of codes has reached 12.

The correct answer is B.

Here are all of the codes that can be formed from A, B, C, D and E:
A
B
C
D
E
AB
AC
AD
AE
BC
BD
BE
CD
CE
DE
Total codes = 15.

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srcc25anu: Master | Next Rank: 500 Posts; Posts: 423; Joined: Fri Jun 11, 2010 7:59 amFollowed by: 2 members; Location: Seattle, WA; Thanked: 86 times

by srcc25anu » Mon Apr 01, 2013 4:19 pm

Lets start with the smallest choice i.e. 4

Single digit codes that can be formed = 4 (a, b, c, d)
Double distinct digit codes iN ALPHABETICAL ORDER out of these 4 letters can be: 4*3/2 = 12/2 codes = 6 CODES
they can be listed as:
AB AC AD BC BD CD
TOTAL CODES = 4 + 6 = 10 (NOT SUFFICIENT)

HENCE WE NEED 5 CODES. THAT SHOULD BE SUFFICIENT TO CREATE A MINIMUM OF 12 CODES
AnSWER b

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Mon Apr 01, 2013 4:40 pm

sanaa.rizwan wrote:A researcher plans to indentify each participant in a certain medical experiment with a code consisting of either a single or a pair of distinct letters written in alphabetical order, what is the least number of letters that can be used if there are 12 participants and each participant is to receive a different code.

A.4
B.5
C.6
D.7
E.8

Another approach is to add a blank to the letters in order to account for the possibility of using just one letter for a code.

For example, consider answer choice A (4 letter).
Let's let the letters be A, B, C, D
We'll also add a "-" to represent a blank.
So, we must choose 2 characters from {A, B, C, D, -}
In how many ways can we select 2 characters?
We can use combinations to answer this. There are 5 characters, and we must select 2. This can be accomplished in 5C2 ways (=10 ways). As you can see, others have already shown those 10 possibilities.

Notice that, when we select 2 characters, there's only 1 possible code that can be created. The reason for this is that the 2 characters must be in alphabetical order. Or, in the case that a letter and a blank is selected, there's only one possible code as well.

Aside: If anyone is interested, we have a free video on calculating combinations (like 5C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Since A is not the correct answer, let's try B (5 letters)
Let's let the letters be A, B, C, D, E
Once again, we'll add a "-" to represent a blank.
So, we must choose 2 characters from {A, B, C, D, E, -}
There are 6 characters, and we must select 2. This can be accomplished in 6C2 ways (= 15 ways).
Perfect. B works.

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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