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Fustrated with combinations and permutation problems!!!

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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Fustrated with combinations and permutation problems!!!

by eclaym2003 » Sat May 24, 2008 8:41 pm
I just bought the Manhattan GMAT book - Word Translations. I am having problems understanding chapter 4 - combinations and permutation. I am having problems with setting up the combinatorics grid. I don't know if to put y's and n's or numbers within this grid depending on the make up of this question.
If anyone has this book or knows of a better way to handle these types of problems I WOULD REALLY APPRECIATE SOME HELP!!!!
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by VP_Tatiana » Tue May 27, 2008 7:52 pm
I would be happy to help. Why don't you throw out a couple examples different types of questions that are stumping you, and I can then write you up a detailed solution. I find this a little more of a concrete way to describe a strategy... than just talking without an example.

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an example of my fustration

by eclaym2003 » Wed May 28, 2008 6:51 pm
I am going to give two examples.

1.) The principal of a high school needs to schedule observations of 6 teachers. She plans to visit one teacher each day of the week, so she will only have time to see 5 of the teachers. How many different observation schedules can she create.

2.) A second grade class is writing reports on birds. The students' teacher has given them a list of four birds they can choose to write about. If Lizzy wants to write a report that includes two or three of the birds, how many different reports can she write?

I am having problems setting up these problems. The books gives a grid to use in order to show the "amount of repeats". Anyway, in the solution to both of these problems the grid is set up differently. Let me know if you need anymore. Thank you sooooo much... I hope this helps!!!
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Re: an example of my fustration

by chipjet » Tue Jun 10, 2008 12:56 pm
eclaym2003 wrote:I am going to give two examples.

1.) The principal of a high school needs to schedule observations of 6 teachers. She plans to visit one teacher each day of the week, so she will only have time to see 5 of the teachers. How many different observation schedules can she create.

2.) A second grade class is writing reports on birds. The students' teacher has given them a list of four birds they can choose to write about. If Lizzy wants to write a report that includes two or three of the birds, how many different reports can she write?

I am having problems setting up these problems. The books gives a grid to use in order to show the "amount of repeats". Anyway, in the solution to both of these problems the grid is set up differently. Let me know if you need anymore. Thank you sooooo much... I hope this helps!!!
1) IMO 720.
Day 1- 6 Options
Day 2- 5 Options
Day 3- 4 Options
Day 4- 3 Options
Day 5- 2 Options

6*5*4*3*2=720

2) IMO 36.
2 Birds:
4 Options
3 Options
4*3=12 combos

3 Birds:
4 Options
3 Options
2 Options
4*3*2=24

24+12=36 possibilities

I'm kind of new to these, too. Are these right?
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by netigen » Tue Jun 10, 2008 2:16 pm
1) When you talk about schedules ordering is important, a different order is a different schedule so this is a Permutation question

Ans = 6P5 = 6!/1! = 6! = 720

2) In this case the order of the choice of birds is not important and hence this is a combination question

ways to choose 2 out of 4 + ways to choose 3 out of 4

= 4C2 + 4C3
= 6 + 4
= 10
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by P_mashru » Thu Jun 12, 2008 3:08 am
2) In this case the order of the choice of birds is not important and hence this is a combination question

ways to choose 2 out of 4 + ways to choose 3 out of 4

= 4C2 + 4C3
= 6 + 4
= 10

There would be multiplication and not addition - b'se she can choose either of combination (3 birds or 2 birds) first and swipe it in between birds

= 4C2 X 4C3
= 6 X 4
= 24

Can anybody authenticate this answer and Diagnose the correct one

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I think that it is starting to make sence

by eclaym2003 » Thu Jun 12, 2008 6:55 pm
When you put it that way it makes sence. I just have problems distinguishing which orders are important an which orders are not.

The answers to the problems are ... the answer to the first problem is 720.
The answer to the second problem is 35.
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by dkrich » Fri Jun 20, 2008 8:20 am
I think the answer to 2 would be 10. You have to figure out two different combinations independently and then add them to get the total number of combinations of either. I believe when "or" is the operative word it is a clue that you need to add, whereas "and" signals multiplication.

The number of unique combinations of 2 birds out of the 4 is 6, and the number of unique combinations of 3 out of 4 is 4, so together there are 10 different possibilities. I believe multiplying those two together instead of adding would give you how many ways those two sets of combinations can be combined which is not what is being asked.
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by Ian Stewart » Fri Jun 20, 2008 9:31 am
netigen and dkrich have the right approaches above.

I don't like the wording of the second question:

"2.) A second grade class is writing reports on birds. The students' teacher has given them a list of four birds they can choose to write about. If Lizzy wants to write a report that includes two or three of the birds, how many different reports can she write? "

No doubt she could write thousands of different reports. The question really is "how many different sets of birds could she choose to write about?"

I like what dkrich says above about seeing 'or' in the question: it's a clue that you'll need to add. The one thing to be sure of is that there's no possible overlap, as we sometimes find in Venn diagram questions. If, at a certain school, 20 people take math, and 30 people take chemistry, that doesn't mean 50 people take math or chemistry, because some might do both- the sets might overlap. Of course, there's no overlap in the birds question above: she can't write about exactly two birds and exactly three birds in the same report. So we can count each case separately and add, to get 10 as was done above.
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