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Question regarding Manhattan Word Problems Book

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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Question regarding Manhattan Word Problems Book

by bpolley00 » Thu Feb 28, 2013 7:19 am
You guessed it, I am going to ask a question in regards to all the GMAT Tutor's favorite, overemphasized problem type: COMBINATIONS/PERMUTATIONS, which if it wasn't for Brian Lange I would have just completely skipped so blame it on him.

Ok so I am going back through Manhattan GMAT's word problem book, and I have a question on Manhattan's explanation of the differences between when to use a simple anagram equation vs. using combinatrics where the individuals are indistinguishable. The book gives 1 example of each:

1) 7 people going on a plane only 3 sets? How many different Arrangements? I understand this is simple angram and no different than 123SSSS. 7/ 4!
2) The second example states if 3 of 7 passengers are selected for a flight, how many different combinations of standby passengers can be selected? Ok again it states combinations so we know that we don't care on the arrangement of bros flying, so it is 7!/4!3! signifying they are coming in combinations or in anagram terms: FFFNNNN

These two are obviously distinguishable as the question specifically states COMBINATION vs. ARRANGEMENT.

NOW what I don't understand is how to distinguish the useage of either 1) or 2) for the multiple arrangement question which is as follows: A frat must choose a delegation of3 senior members and two Jr. members for an annual interfrat conference. IT has 12 Sr. and 11 Jr, how many different delegations are possible?

So you are just assuming that it is asking for a combination here rather than an anagram? I mean I know you aren't assuming anything, there must be a way to distingusih this. How did you make that distinction as it is vitally important on the test - if I see one of these God awful question I want to get it right.

If you could please elaborate that would be much appreciated. Thanks in advance.

-BP
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by Anurag@Gurome » Tue Mar 05, 2013 1:13 am
bpolley00 wrote:... I mean I know you aren't assuming anything, there must be a way to distingusih this. How did you make that distinction as it is vitally important on the test - if I see one of these God awful question I want to get it right.
Selections and arrangements are different things.
When you have to arrange something, first you have to select them.

For example, say there are 5 people but 3 chairs. In how many ways the 5 people can sit on the 3 chairs?

First we have to select 3 people from 5 people ---> Selection
In how many ways we can do that? ---> 5C3 = 5!/[(2!)*(3!)] = 10

Now, for each of the above selection, we have to arrange this 3 people in three chairs ---> Arrangement
In how many ways we can do that? ---> 3! = 6 ways

hence, total number of arrangements = [5!/[(2!)*(3!)]]*(3!) = 5!/2! = 60

In mathematical terminology,
  • Selection == Combination (Formula --> nCr = n!/[((n - r)!)*(r!)]
    Arrangement = Permutation (Formula --> nPr = n!/[(n - r)!]
So the trick is to understand whether the problem is asking for number of selections or arrangements? If it is asking for number of selections only, we stop after selection. But if it is asking for number of arrangements, first we will select then arrange.

For example, the question you chose,
bpolley00 wrote: A fraternity must choose a delegation of 3 senior members and 2 junior members for an annual conference. It has 12 senior and 11 junior candidates, how many different delegations are possible?
It is obvious that we need to select 3 from 12 seniors and 2 from 11 juniors. But after that do we need to arrange them? No. Why not? Because this is a delegation, a simple team forming. For example, if A, B, and C are three seniors and X and Y are two juniors, ABCXY/CABYX/BXACY etc all are same team. Therefore, we don't need number of arrangements. Calculating number of selections is sufficient.

Hence, number of different delegations = (Number of ways to select 3 from 12)*(Number of ways to select 2 from 11) = (12C3)*(11C2)

But what if the question said that in a delegation there will be a post of representative or leader and only a senior can occupy that place?

Now, in a delegation there will be 3 seniors and 2 juniors with one senior occupying the post of leader. This means for each selection we made earlier, the delegation will change depending upon which one of the three seniors is the leader.

If we assume the first position is for leader, then ACBXY and CABXY are different delegations. But ACBXY/ACXBY/ABCYX etc are still same delegation as the other posts are inter-changeable.

Now, number of delegations = (Number of ways to select 3 from 12)*(Number of ways to select 2 from 11)*(Number of ways to arrange 3 seniors among themselves) = (12C3)*(11C2)*(3!) = (12P3)*(11C2)
Anurag Mairal, Ph.D., MBA
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by bpolley00 » Tue Mar 05, 2013 2:31 pm
Dr. Mairel,

That explanation was crystal clear, much more so than the Manhattan Books. Thanks so much for your poignant response sir.
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