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A group of candidates for two analyst positions consists of six people

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A group of candidates for two analyst positions consists of six people  [#permalink]

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New post 08 Apr 2015, 04:59
4
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A
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D
E

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A group of candidates for two analyst positions consists of six people. If one-third of the candidates are disqualified and three new candidates are recruited to replace them, the number of ways in which the two job offers can be allocated will:

(A) Drop by 40%
(B) Remain unchanged
(C) Increase by 20%
(D) Increase by 40%
(E) Increase by 60%


Kudos for a correct solution.

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Re: A group of candidates for two analyst positions consists of six people  [#permalink]

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New post 08 Apr 2015, 05:09
1
2
Initially 6 people for 2 positions, meaning
6 - 1st position
5 - 2nd position

Total combinations- 5*6 = 30

Now 1/3rd not qualified, so reduce by 2. Hence, now we have 4+3 candidates for the position.

Same as above, 7*6 for two positions will give 42 combinations. Thus, 42-30/30 = 40% increase

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A group of candidates for two analyst positions consists of six people  [#permalink]

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New post 08 Apr 2015, 05:43
First scenario: 2 people to pick from 6 candidates. Since we don't take into account order, the formula is \(6C2\) which is 15 ways
Second scenario: 2 people to pick from 7 candidates. Same story, \(7C2\) which is 21 ways
\(\frac{(21-15)}{15} = 0,4 = 40%\) = D
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Re: A group of candidates for two analyst positions consists of six people  [#permalink]

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New post 08 Apr 2015, 06:51
Bunuel wrote:
A group of candidates for two analyst positions consists of six people. If one-third of the candidates are disqualified and three new candidates are recruited to replace them, the number of ways in which the two job offers can be allocated will:

(A) Drop by 40%
(B) Remain unchanged
(C) Increase by 20%
(D) Increase by 40%
(E) Increase by 60%


Kudos for a correct solution.



case 1: \(6C2\)= 15 ways
case 2: \(7C2\) = 21 ways
\(\frac{(21-15)}{15} = 40%\)

answer D
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Re: A group of candidates for two analyst positions consists of six people  [#permalink]

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New post 08 Apr 2015, 13:32
1
Total 6 candidate 2 job positions; so # of ways position can be arranged = 6C2 =15
Then 1/3 of 6 disqualified=2, and 3 new added=6-2+3=7 total # of candidate and 2 job positions; # of ways position can be allocated=7C2=21
Hence change = ((21-15)/15)*100=40 %, increase

Hence answer is D

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Re: A group of candidates for two analyst positions consists of six people  [#permalink]

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New post 13 Apr 2015, 06:26
Bunuel wrote:
A group of candidates for two analyst positions consists of six people. If one-third of the candidates are disqualified and three new candidates are recruited to replace them, the number of ways in which the two job offers can be allocated will:

(A) Drop by 40%
(B) Remain unchanged
(C) Increase by 20%
(D) Increase by 40%
(E) Increase by 60%


Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

After reading such a question, you may still not be sure what to do, but you can start piecing together the issue at hand. There are six people interviewing for two jobs, but then some will drop out and others will join, and the overall impact must be gauged. The answer choices seem to offer various increases and decreases, so the answer must be in terms of the adjustment of job offer possibilities. This makes the question seem like a combinatorics or probability question.

Looking at the information provided, we have six applicants for two positions, and then one-third of them are disqualified. This leaves us with four finalists for the two jobs (like musical chairs), but before a decision is rendered, three more applicants join. There are now seven candidates for the two jobs, yielding a net change of one new contender. From 6 to 7 people, the change would be 1/6 of the old total, or 16.7%. This is closest to answer choice C, but there is no direct match among the answer choices. Since the GMAT doesn’t provide horseshoe answer choices (unless approximation is specified), this is our first hint that we may need to dig deeper in our approach.

The questions specifically asks about “the number of ways in which the two job offers can be allocated”, which should hopefully make you realize that the question is ultimately about permutations. In the initial setup, two positions are available for six candidates, meaning we can calculate the number of possible outcomes.

The only decision we have to make is about the order mattering, and since it’s not indicated anywhere that the jobs are identical, it’s reasonable to assume we can differentiate between job 1 and job 2. Let’s say that the first job is a senior position and the second is a junior position, how many ways can we fill these openings? Anyone can take the first position, so that gives us 6 possibilities, and then anyone of the remaining choices can fill the second position, yielding another 5 possibilities. Since any of these can be combined, we get 6*5 or 30 choices. Using the permutation formula of N!/(N-K)! yields 6! /4! which is still 6*5 or 30, confirming our answer.

If there were 30 possibilities at first, the addition of a new candidate will undoubtedly increase the number of possibilities, so we can consider answer choices A and B eliminated. After the increase, we can essentially make the same calculations for 7 candidates and 2 jobs, giving us 7*6 or 42 choices. We used to have 30 choices and now we have 42, so that works out to 12 new choices out of the original 30, equivalent to a 40% increase. Answer choice D is a 40% increase, and thus exactly the correct answer.

Some of you may be asking about the assumption I made about order mattering a few paragraphs back. “Ron, Ron”, you ask, “what happens if we assume that the order doesn’t matter?” Let’s run the calculations to see. If the order doesn’t matter and we’re dealing with a combination, then we have 6 candidates for 2 positions, we will get N! / K! (N-K)! which is 6! / 2! * 4! Simplifying to 6*5 / 2 gives us 15 options instead of the previous 30. Really, these are the same options but now we divide by two because the order no longer matters (i.e. AB and BA are equivalent). The updated scenario will have 7! / 2! * 5!, which becomes 42 / 2 or 21. This is exactly half the previous number again. The delta from 15 to 21 is 6, again 40% of the initial sum of 15. Since we’re dealing with percentages, both combinations and permutations will be completely equivalent. (Ain’t math grand?)

Regardless of minor assumptions made while solving this problem, the solution will always be the same. Indeed, the hardest part of solving the problem is often determining what is being asked. Remember that there can only be one answer to the problem, and that the answer choices can help steer you in the right direction. If you know what you’re looking for, the questions on the GMAT may be somewhat vague, but your goal will be crystal clear.
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Re: A group of candidates for two analyst positions consists of six people  [#permalink]

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New post 15 Apr 2016, 04:44
Hey @vyasak everyone is writing that two jobs can be filled in 15 and 21 ways
but should we be applying permutation here .
here a,b will be different from b,a
so the number of ways should be 30,42

Your call?
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Re: A group of candidates for two analyst positions consists of six people  [#permalink]

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New post 15 Apr 2016, 10:45
stonecold wrote:
Hey @vyasak everyone is writing that two jobs can be filled in 15 and 21 ways
but should we be applying permutation here .
here a,b will be different from b,a
so the number of ways should be 30,42

Your call?


No we must apply the concept of Combination since the problem is dealing with selection.

Here is my approach -

people = 6
disq = 2
Left = 4
Added = 3

total = 7

6C2 = 15
7C2 = 21

Incr = 6

% increase is 6/15 * 100 =40% increase
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Re: A group of candidates for two analyst positions consists of six people  [#permalink]

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New post 16 Apr 2016, 06:50
Abhishek009 wrote:
stonecold wrote:
Hey @vyasak everyone is writing that two jobs can be filled in 15 and 21 ways
but should we be applying permutation here .
here a,b will be different from b,a
so the number of ways should be 30,42

Your call?


No we must apply the concept of Combination since the problem is dealing with selection.

Here is my approach -

people = 6
disq = 2
Left = 4
Added = 3

total = 7

6C2 = 15
7C2 = 21

Incr = 6

% increase is 6/15 * 100 =40% increase



I think we need permutation here bro.
Exact same question in RD sharma 11 class
CC=> Vyshak
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A group of candidates for two analyst positions consists of six people  [#permalink]

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New post 16 Apr 2016, 07:05
stonecold wrote:
Abhishek009 wrote:
stonecold wrote:
Hey @vyasak everyone is writing that two jobs can be filled in 15 and 21 ways
but should we be applying permutation here .
here a,b will be different from b,a
so the number of ways should be 30,42

Your call?


No we must apply the concept of Combination since the problem is dealing with selection.

Here is my approach -

people = 6
disq = 2
Left = 4
Added = 3

total = 7

6C2 = 15
7C2 = 21

Incr = 6

% increase is 6/15 * 100 =40% increase



I think we need permutation here bro.
Exact same question in RD sharma 11 class
CC=> Vyshak


Hi,
If the Q talks of TWO analysts position, there is no reason WHY should I take it as different positions as suggested by the Source-- ONE junior position and the OTHER senior position..
I do not think we should work on assumptions and why should we? Combinations is perfectly fine here, there is no ambiguity APART from a word 'allocated'...

Although either way answer will be same ..
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Re: A group of candidates for two analyst positions consists of six people  [#permalink]

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New post 16 Apr 2016, 09:14
stonecold wrote:
Hey @vyasak everyone is writing that two jobs can be filled in 15 and 21 ways
but should we be applying permutation here .
here a,b will be different from b,a
so the number of ways should be 30,42

Your call?


Hi stonecold,

Chetan is right. The question does not differentiate between the 2 positions.

We have to apply permutation when the question asks us to select and arrange. Here the question just asks us to select 2 candidates for the post of the analyst. The arrangement does not matter.
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Re: A group of candidates for two analyst positions consists of six people  [#permalink]

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New post 07 Jun 2017, 21:26
Bunuel wrote:
A group of candidates for two analyst positions consists of six people. If one-third of the candidates are disqualified and three new candidates are recruited to replace them, the number of ways in which the two job offers can be allocated will:

(A) Drop by 40%
(B) Remain unchanged
(C) Increase by 20%
(D) Increase by 40%
(E) Increase by 60%


Kudos for a correct solution.


What this question is basically asking is what is the percentage increase from the original amount of total combinations to the new amount of total combinations; the total combinations of candidates from a pool that comprises the original qualified candidates and the new qualified candidates.

6(1/3) = 2 disqualified
6(2/3) = 4 qualified ( you don't actually have to calculate this- this is just shown for the sake of understanding the problem)

6c2 = 15
7c2= 21

15(x) = 21

Now, it is important to remember you can do a percentage increase by converting the percentage and multiplying it ( a 20% increase of x is x(1.2) - or you could do 210/15 which would also give you 40 percent instead of testing answer choices repeatedly.
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Re: A group of candidates for two analyst positions consists of six people  [#permalink]

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New post 08 Feb 2018, 16:51
Bunuel wrote:
A group of candidates for two analyst positions consists of six people. If one-third of the candidates are disqualified and three new candidates are recruited to replace them, the number of ways in which the two job offers can be allocated will:

(A) Drop by 40%
(B) Remain unchanged
(C) Increase by 20%
(D) Increase by 40%
(E) Increase by 60%



The number of ways to choose 2 candidates from 6 is 6C2 = 6![2!(6-2)!] = 6!/(2! 4!) = (6 x 5)/2! = 30/2 = 15. After 2 of the 6 candidates leave and 3 new candidates are recruited, there will be 7 candidates, and the number of ways to choose 2 candidates from 7 is 7C2 = 7!/[2!(7-2)!] = 7!/(2!5!) = (7 x 6)/2! = 42/2 = 21. Thus, the number of ways to choose 2 candidates is increased by

(21 - 15)/15 x 100% = 6/15 x 100% = 2/5 x 100% = 40%

Answer: D
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