Problem Solving

A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair or a pair of distinct letters in alphabetic 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

I dont have the OA, but the answer given in the doc file from where I picked this up is B

Here is my answer to such a question but I am not sure if it is correct: -

Partcipant #1= a
Partcipant #2= aa
Partcipant #3= b
Partcipant #4= ab
Partcipant #5= bb
Partcipant #6= c
Partcipant #7= ac
Partcipant #8= bc
Partcipant #9= cc
Partcipant #10= d
Partcipant #11= ad
Partcipant #12= bd

The letters used were: a, b, c, & d.......4 so my answer is [A] I would appreciate it if someone confirms the approach.

THE_BOSS wrote:Here is my answer to such a question but I am not sure if it is correct: -

Partcipant #1= a
Partcipant #2= aa
Partcipant #3= b
Partcipant #4= ab
Partcipant #5= bb
Partcipant #6= c
Partcipant #7= ac
Partcipant #8= bc
Partcipant #9= cc
Partcipant #10= d
Partcipant #11= ad
Partcipant #12= bd

The letters used were: a, b, c, & d.......4 so my answer is [A] I would appreciate it if someone confirms the approach.

Very close.....The stem states that "either a single letter or a pair of DISTINCT letters in alphabetic order" Since they have to be disctinct...you cannot have AA, BB, CC and DD...

So,
A
B
C
D
AB
AC
AD
BC
BD
CD

so the 4 albhabets only give us 10...if we add to E to the mix...we can get 5 more...which would be sufficient for the 12 participants..

----
E
AE
BE
CE
DE
----

Hence over all we would need 5 alphabets..so the answet is B.

Cheers.

Is there a formula that we could use for this kind of questions?

@ THE_BOSS

It would be advisable if you go with making combinations with such questions. That would be an easier approach. Hardly took 30 secs.
start off with the lowest option i.e with 4 , make combinations and see how close you are to the answer.
However, I solved with the following techinque.
Taking n =4 , no of single digit codes = 4
and no of 2 digit code = 4C2 = 6
Total no of codes = 10.
so, this gives me an idea that 5 (next number will satisfy the condn)would be the number that will give me atleast 12 different codes.
Hence B

ankitns wrote:

THE_BOSS wrote:Here is my answer to such a question but I am not sure if it is correct: -

Partcipant #1= a
Partcipant #2= aa
Partcipant #3= b
Partcipant #4= ab
Partcipant #5= bb
Partcipant #6= c
Partcipant #7= ac
Partcipant #8= bc
Partcipant #9= cc
Partcipant #10= d
Partcipant #11= ad
Partcipant #12= bd

The letters used were: a, b, c, & d.......4 so my answer is [A] I would appreciate it if someone confirms the approach.

Very close.....The stem states that "either a single letter or a pair of DISTINCT letters in alphabetic order" Since they have to be disctinct...you cannot have AA, BB, CC and DD...

So,
A
B
C
D
AB
AC
AD
BC
BD
CD

so the 4 albhabets only give us 10...if we add to E to the mix...we can get 5 more...which would be sufficient for the 12 participants..

----
E
AE
BE
CE
DE
----

Hence over all we would need 5 alphabets..so the answet is B.

Cheers.

what does this mean "a single letter or a pair or a pair of distinct letters" . Doesnt this say that you can even have a pair of non-distinct letter ?

Hmm...i think "or a pair or a pair of.." is a typo and should just be "or a pair of.."

Can someone confirm?

Thanks.

rish wrote:

ankitns wrote:

THE_BOSS wrote:Here is my answer to such a question but I am not sure if it is correct: -

Partcipant #1= a
Partcipant #2= aa
Partcipant #3= b
Partcipant #4= ab
Partcipant #5= bb
Partcipant #6= c
Partcipant #7= ac
Partcipant #8= bc
Partcipant #9= cc
Partcipant #10= d
Partcipant #11= ad
Partcipant #12= bd

The letters used were: a, b, c, & d.......4 so my answer is [A] I would appreciate it if someone confirms the approach.

Very close.....The stem states that "either a single letter or a pair of DISTINCT letters in alphabetic order" Since they have to be disctinct...you cannot have AA, BB, CC and DD...

So,
A
B
C
D
AB
AC
AD
BC
BD
CD

so the 4 albhabets only give us 10...if we add to E to the mix...we can get 5 more...which would be sufficient for the 12 participants..

----
E
AE
BE
CE
DE
----

Hence over all we would need 5 alphabets..so the answet is B.

Cheers.

what does this mean "a single letter or a pair or a pair of distinct letters" . Doesnt this say that you can even have a pair of non-distinct letter ?