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Permutations & Combinations : Help is on the way! [#permalink]
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13 Oct 2004, 22:53
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I personally do not think the permutations and combinations formulas are necessary to do well on the test, and I think seeing them can be very scary and daunting.
While I acknowledge that there are an infinate number of variations on how these questions could be asked, I think for the most part the test is not testing our ability to use the formulas but rather our ability to understand what's going on behind them.
Therefore, they ask questions that the formulas won't work with, but logic will.
For example: How many ways can five people sit in a five person car if only 2 can drive?
There's no formula for that  it's just 2x4x3x2x1.
So in my opinion (and this is how I teach my students) it's more important to learn why that works and how to structure the numbers, and why to divide by whatever you need to divide by in a combinations question.
Then there's nothing to fear.



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ian7777 wrote: I personally do not think the permutations and combinations formulas are necessary to do well on the test, and I think seeing them can be very scary and daunting.
While I acknowledge that there are an infinate number of variations on how these questions could be asked, I think for the most part the test is not testing our ability to use the formulas but rather our ability to understand what's going on behind them.
Therefore, they ask questions that the formulas won't work with, but logic will.
For example: How many ways can five people sit in a five person car if only 2 can drive?
There's no formula for that  it's just 2x4x3x2x1.
So in my opinion (and this is how I teach my students) it's more important to learn why that works and how to structure the numbers, and why to divide by whatever you need to divide by in a combinations question.
Then there's nothing to fear.
Ian7777, can you pls explain how the answer is 2 x4x3x2x1? Thanks



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Folaa3 wrote: ian7777 wrote: I personally do not think the permutations and combinations formulas are necessary to do well on the test, and I think seeing them can be very scary and daunting.
While I acknowledge that there are an infinate number of variations on how these questions could be asked, I think for the most part the test is not testing our ability to use the formulas but rather our ability to understand what's going on behind them.
Therefore, they ask questions that the formulas won't work with, but logic will.
For example: How many ways can five people sit in a five person car if only 2 can drive?
There's no formula for that  it's just 2x4x3x2x1.
So in my opinion (and this is how I teach my students) it's more important to learn why that works and how to structure the numbers, and why to divide by whatever you need to divide by in a combinations question.
Then there's nothing to fear. Ian7777, can you pls explain how the answer is 2 x4x3x2x1? Thanks
Yes. There are 5 seats in the car. You need only ask yourself logically how many options there are for each seat. The first seat is the drivers seat, so there are only 2 who can sit there. Once one person sits down, however, the remaining 4 are available for the second seat. Then there are 3, and then 2, and then 1. We just multiply that all together for the answer.



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I dont mean to be presumptuous here but there is a formula for that. Yes it needs to be developed from logic definitely. WHat i felt the formula in this case is
out of n r are of one type and nr are the other type. while r can replace nr but nr cannot replace r .
Now it reamins to be seen how mnay places the kind "r" alone needs to occupy. That will help to formulate. Lets call it k ( r>k) . Then the places for nr objects of one kind and r objects of the other kind can be arranged as
[ nr (rk) ] ! * r!
to test ,n = 5 r =2 k=1
nr (rk) = 4
4! is the number of ways the passengers can be seated. there are 2 drivers , therefore the number of ways is 4!*2!
It can b chekced with similar arrngements . seems to work with circular permutations too.
Ok before i ruffle anyone completely . I did this purely for academic reasons. I know it is lengthy and screwy.. but yeah if i can develop a habit of finding patterns early, i may not have to actually write down the formula , but can bring out the logic really fast. Once again not trying to show off!!!!



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ok so say you dont want to use the formulas can u give an example of when something might be a permutation rather than a combination and the logic for figuring that out vs. the logic for figuring out a combination?



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Rupstar wrote: ok so say you dont want to use the formulas can u give an example of when something might be a permutation rather than a combination and the logic for figuring that out vs. the logic for figuring out a combination?
yes. spend some time with the concepts of permutations and combinations and learn why one is different from the other. You'll see that combinations are just permutations with an extra division. So go back further and learn why permutations work. They work because when we multiply all the options together, simply, we get the total number of outcomes. So be logical and simply figure out what the options are for each piece of the arrangement, and you'll have the answer. If the order doesn't matter, just divide that answer by the number of spaces factorial. If you're not sure why, learn why. It's because you're canceling out all the repeated arrangements and are just left with the distinct groups.
When you know why these things work, you have power, knowledge, and an edge over this test. When you just learn formulae, you have the potential to make a really big mistake.



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yep, probably the formulas are not necessary after all but knowing how to solve perm and comb problems is needed.
I commit the mistake of thinking that I was not going to need it to get 600+; however, on test day the fifth question was about perm and comb



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Imagine you order a pizza, and you want a three topping pizza. You will have pepperoni, mushrooms, and onions. The waiter asks you, "which topping do you want on your pizza first?"
What would you answer? Would you ever expect this question? No. Because the order of the toppings wouldn't matter. Your pizza will have three toppings, end of story.
Now, if you did want to know how many ways you could put those three toppings on your pizza, you could figure it out with permutations: 3 x 2 x 1 = 6. But at the end of the day, they're all the same, so you don't count them.
What would be different? Well, a different pizza, that's what: sausage, pinapple, and onion would be a different pizza. Again, you wouldn't care about the order, but it would be different.
That's the difference between permutations and combinations. In permutations, we count all the orderings of every different group. In that case, the two pizza examples would be 12 different permutations. But in combos, we don't care about the order, so we neutralize the copies. Take that 12 permutations from above, divide by the number of copies of each group, and that's combos, in this case: 2.
So combinations is a permutations with a tax on it. You manipulate a permutation question to get a combinations answer. And you do it by dividing by the number of spaces in the permutation, factorial.
And that's how it goes. Look for examples of each on the web  the common divisor between all combos is that we won't care about the order.



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Permutation
Definition: An arrangement of objects whose order is important.
Example: Permutation of objects A B C results
ABC, ACB, BAC, ACA, CAB, CBA
There are 6 orders, or 6 permutations in total
Factorial:
Example: The total permutations on ABC is
3!=3 X 2 X 1=6



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Example 2: To find all permutations on A B C C
Answer: ABCC, ACBC, ACCB,
BACC, BCAC, BCCA,
CABC, CBAC, CBCA,
CACB, CCAB, CCBA
There are 12 orders, or 12 permutations in total.
P( permutation of n objects with n1 identical ones)
= n!/ n1!
Example 2 of ABCC:
n = 4; (A,B,C,C)
n1= 2; (C,C)
n!/ n1! = 4! / 2! = (4 x 3 x 2 x 1)/ (2 x 1) = 12



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Combination question baffling my mind [#permalink]
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01 Jul 2006, 17:52
Came across this question on this site itself and I am struggling to get the answer the way they describe it and why my approach is at fault
here is the question:
EXAMPLE 10 A threeperson committee must be chosen from a group of 7 professors and 10 graduate students. If at least one of the people on the committee must be a professor, how many different groups of people could be chosen for the committee?
A. 70 B. 560 C. 630 D. 1,260 E. 1,980
My approach :
Since we need to have a professor for the 3member committee , we can choose the first person as professor in 7 ways . Rest two positions can be filled with remaining professors or students in (16 C 2 )
total number of combinations for filling 3 member committe  first position with professor * rest with 2 from 16 mombers ( 6 professors + 10 students) is 7 * (16 C 2) = 7 * (16 * 15 )/2 = 840
but what is baffling is total number of combinations for choosing 3 members from 17 ( 7 professors + 10 students) without restriction of atleast one professor in the committie = 17 C 2 = 680
how can this combination be less than restricted combination
can someone explain me where my thought process took a wrong turn ..



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This is a great question. Your problem is a very common one. I've seen it a lot. I hope this discussion will help a lot of people to clear things up.
In your approach, you first chose 1 professor. Then you chose the rest two people from the rest 16 people. The problem is that you would have some repeats here. For example, you may have chosen P1, and then P2 and S1. But it is also possible that you would have chosen P2, and then P1 and S1. In your approach these two cases are counted as two outcomes. However if you look at them, it's the same group of three people. What you should have done, is to break it into three cases, where 1, 2, or 3 professor(s) would be included, so that you know for sure you will not be double counting.
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I am also in dire need of help with this one:
In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed?
(A) 5
(B) 4
(C) 3
(D) 2
(E) 0



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amartin6165 wrote: I am also in dire need of help with this one:
In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed? (A) 5 (B) 4 (C) 3 (D) 2 (E) 0 Answer is D. See below link http://gmatclub.com/phpbb/viewtopic.php?t=31621
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P&C in simple way.. [#permalink]
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21 Nov 2006, 09:50
====================================
Came across this question on this site itself and I am struggling to get the answer the way they describe it and why my approach is at fault
here is the question:
EXAMPLE 10 A threeperson committee must be chosen from a group of 7 professors and 10 graduate students. If at least one of the people on the committee must be a professor, how many different groups of people could be chosen for the committee?
A. 70
B. 560
C. 630
D. 1,260
E. 1,980
=============================================
BLISSFUL wrote:
Easiest way to solve the above Q is:
Total ways of choosing 3 from 7 profs & 10 students = 17C3
Total ways in which there is NO Prof = Total ways in which ONLY students are there = 10C3.
Now Total ways in which ATLEAST ONE prof is there is:
Total possible ways of choosing 3 from 17  Total ways of choosing all 3 as students( In all other selections, there will be ATLEAST one prof present )= 17C310C3 = 560 Ans B



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I arrived at the correct answer but when trying the alternative method I cant understand why it doesn't work:
Number possibilities for 1st member = 7
Number possibilities for 2nd member = 16
Number possibilities for 3rd member = 15
So why not 7*16*15/3*2*1 = 260????????
It's getting late  I need to sleep!



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Re: Permutations & Combinations : Help is on the way! [#permalink]
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22 Mar 2008, 12:32
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Here is an example to something similar I saw in one of the OG or Kaplan books that I can show the difference between a combination and a permutation. Two versions of the question. Permutation Version:There are 11 chemicals jars in a laboratory that are identified by a two color labeling system. How many colors are needed to identify all 11 chemicals? The answer is: 4 colorsQuick solution.... 4 labels using the same color twice 114= 7 labels remaining that will use two different colors 4*3 = 12 additional color combinations so this is sufficient Longer solution... You know you will need significantly less then 11 colors to do this so pick a number like 3 to start with. three colors(R, G, and B) will give you three labels with: RR, GG, and BB You still need 113 = 8 labels. The resulting possibilities of mixing the three colors are: RG, RB, GR, GB, BR, and BG = 6 labels 3+8 = 11. Not enough possibilites. So, 3 is barely too small. Work up from there. Let's try 4. four colors(R, G, B, and Y) will give you four labels with: RR, GG, BB, and YY You still need 114 = 7 labels. The resulting possibilities of mixing the four colors are: RG, RB, RY, GR, GB, GY, BR, BG, BY, YR, YG, and YB = 12 labels 4+12 = 16 label possibilities which is sufficient. Combination Version:There are 11 chemicals jars in a laboratory that are identified by a two color labeling system. How many colors are needed to identify all 11 chemicals if the order on the label doesn't matter? The answer is: 5 colorsQuick solution.... 5 labels using the same color twice 115= 6 labels remaining that will use two different colors (5*4)/2 = 10 additional color combinations. 10+5 = 15 total possibilities with 5 colors. Longer solution... The same approach applies so we will start out with 4 colors. four colors(R, G, B, and Y) will give you four labels with: RR, GG, BB, and YY You still need 114 = 7 labels remaining The resulting possibilities of mixing the four colors are: RG, RB, RY, GR, GB, GY, BR, BG, BY, YR, YG, and YB = 12 labels Since the problem said that color order on the label doesn't matter so RG is the same as GR. We have to eliminate the duplicates. 12/2 = 6 possibilities. 4+6 = 10, which is not sufficient. Let's try five colors (R, G, B, Y, and O). These will give you: RR, GG, BB, YY, and OO = 5 labels. 115 =6 labels remaining. The resulting possibilities of mixing the four colors are: RG, RB, RY, RO, GR, GB, GY, GO, BR, BG, BY, BO, YR, YG, YB, YO, OR, OG, OB, and OY = 20 possibilities. Again, since the problem said that color order on the label doesn't matter we have to eliminate the duplicates. 20/2 = 10. 10+5 = 15 total label possibilites from 5 colors which is sufficient. Another example: Permutation Version:There are 4 teams in a league. How many different games are possible?4*3 = 12 games possible. 12, 13, 14, 21, 23, 24, 31, 32, 34, 41, 42, 43 Combination Version: There are 4 teams in a league. How many different games are possible if two teams only play each other once?(4*2)/2 = 6 games possible. 12, 13, 14, 23, 24, 34 For combinations you are just getting rid of duplicates. Hope this helps.




Re: Permutations & Combinations : Help is on the way!
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