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2 problems... solve plz

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2 problems... solve plz

by myornob » Tue Sep 04, 2007 10:10 pm
1) A researcher plan to identify each participant in a certain medical experiment with a code consisting of either a single letter 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 participants is to receive a different code.

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

Ans: (B); but I guess the answer is C

2) How many different 6-letter sequences are there that consist of 1 A, 2 B's, and 3 C's

A) 8%
B) 15%
C) 45%
D) 52%
E) 56%

Ans: C-(45%)

Plz. solve this two problem
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by samirpandeyit62 » Wed Sep 05, 2007 12:04 am
1st problem:

As per Q we assign a unique single lettler or a distinct pair of letters in alphabetical Order to 12 partcipants

Now here we can visualise that if we use single lettlers onlt they we will need 12 letters, however if we use letter pairs, then we can minimalise the nos of letters used, by assigning pairs and then the letters individually
to sum up we need to emphasise on assigning letter pairs.

i.e our assignment should be 1st assign letters pairs to as many as possible and then assign individual letters to remaining.

Hence we need to find the highest combination of 2 letters whose value is < 12
i.e. try out 4c2 which gives 6 OK
5c2 gives 10 ok
6c2 gives 15 STOP (only 12 codes are required)

hence the highest Combination of 2 letters whose value is less that 12 is 10

Hence the answer is B
(Intrepret as " We can use 5 letters to assign 10 letter pair codes & 2 single letter codes to 12 partcipants)
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Thanks samir

by myornob » Wed Sep 05, 2007 5:25 am
Oh God... I've taken single Letter code first then have consider the order pair sequence.

So, it'll be actually opposite (consider pair first) to make a least alphabet choice.

But, my second question......? anyone.
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by samirpandeyit62 » Wed Sep 05, 2007 6:00 am
In the second problem can u pls specify percent of what is required
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actual answer choices

by myornob » Wed Sep 05, 2007 8:30 am
Sorry bro, I've put up wrong answer choices.......mistakenly done..

Q. How many different 6-letter sequences are there that consists of 1 A, 2 B's ans 3 C's?

But the answer choices are-

A) 6
B) 60
C) 120
D) 360
E) 720
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by samirpandeyit62 » Wed Sep 05, 2007 9:13 am
We need to arrange 1 A 2 B's & 3 C's i.e. total 6 letters,

now if there are no constraints like both B's must be together etc

this should just be a clean permutation i.e 6P6 = 6! =720

so ans should be E

Pls check if this is correct
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by jaspreet.sharma » Wed Sep 05, 2007 9:38 am
samirpandeyit62 wrote:We need to arrange 1 A 2 B's & 3 C's i.e. total 6 letters,

now if there are no constraints like both B's must be together etc

this should just be a clean permutation i.e 6P6 = 6! =720

so ans should be E

Pls check if this is correct
Shouldnt the answer be B??

as we have 2 b and 3 c

the answer should be.....

6!/(2!*3!) = 60

correct me if am wrong.....
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by myornob » Wed Sep 05, 2007 9:57 am
Yes, the correct answer is B (60)
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by gabriel » Wed Sep 05, 2007 10:06 am
..@ myornob .. please post each question in a new thread .. will help everyone keep track of the questions .. thanx
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by optimisticsam » Wed Sep 05, 2007 12:00 pm
On first question - 2 letter codes must be in alphabetical order according to Q.

So therefore we can't do straight permutation, can we?

Should it not be 7 letters?

A,B,C,D,E,F,G and AB, BC, CD, DE, EF = 12 combinations?

If I am way off base, please explain?

Thanks. (kind of new at this)
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by samirpandeyit62 » Wed Sep 05, 2007 10:07 pm
Hi All,
Well didn't give any time to this one, Yes Jaspreet u are correct, the answer should be 60, the reason is that in these type of problems, the repeated letters do not make any permutation among themselves i.e

if there are 2 repeated letters they would not make a permutation of 2! unlike different letters, irrespective of whether they are placed adjacent to each other or seperated by other letters.

so in such problems we need to divide the overall permutaion of the letters with the factorials of repeated letters equivalent to nos of times they are repeated.

This can also be solved using Combinations as here Order doe'st exist for repeated letters :

A can occupy 6 positions in 6C1 = 6 ways
2 B's can occupy remaining 5 positions in 5C2 (We use Combination as Order does't exist here)= 10 ways
Remaining 3 C's can ocuupy the 3 left positions in 3C3 =1 way

hence altogether it is 6X10X1 =60 ways

But I would suggest that everyone use the first Approach i.e the one Jaspreet used as this one may involve use of both permutation & combinations in certain problems.
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by samirpandeyit62 » Wed Sep 05, 2007 11:07 pm
Hi optimisticsam,
You are correct, but if u see my solution, I have used Combinations instead of Permutations

as suppose 2 letters are A, B so you can use only AB and not BA, which would be taken care by Combinations as they would pick the codes only in one Order. (Alphabetically also imposes this only, pick up in one order Order i.e AB not BA)
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by tanyajoseph » Tue Sep 11, 2007 9:23 am
For the first question, it asks for least alphabets that can be used.

Should it not be 4???

A,B,C,D,AB,BA,BC,CB,CD,DC,AC,CA - distant alphabets and codes.
Could sme1 pls explain why this would be wrong?

Thanks
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