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Amy and Adam are making boxes of truffles to give out as wed

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Amy and Adam are making boxes of truffles to give out as wed  [#permalink]

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New post 20 Jun 2011, 19:16
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Amy and Adam are making boxes of truffles to give out as wedding favors. They have an unlimited supply of 5 different types of truffles. If each box holds 2 truffles of different types, how many different boxes can they make?

I understand these are all the possible combinations:

Box 1: A B
Box 2: A C
Box 3: A D
Box 4: A E
Box 5: B C
Box 6: B D
Box 7: B E
Box 8: C D
Box 9: C E
Box 10: D E

But I don't get the mathematical interpretation of this.. 5! / (3! * 2!)

I don't get how the anagram can be "YYNNN"
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Re: MGMAT Word Translations COMBINATORICS!!!  [#permalink]

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New post 21 Jun 2011, 01:39
Mahtab wrote:
Amy and Adam are making boxes of truffles to give out as wedding favors. They have an unlimited supply of 5 different types of truffles. If each box holds 2 truffles of different types, how many different boxes can they make?

I understand these are all the possible combinations:

Box 1: A B
Box 2: A C
Box 3: A D
Box 4: A E
Box 5: B C
Box 6: B D
Box 7: B E
Box 8: C D
Box 9: C E
Box 10: D E

But I don't get the mathematical interpretation of this.. 5! / (3! * 2!)

I don't get how the anagram can be "YYNNN"


You can think of it in 2 ways:
1. You have 5 different types of truffles. For each box, you have to select 2 different types of truffles. In how many ways can you do it?
5C2 = 5!/3!*2!
(When out of n things, you have to select r, you can do it in nCr ways. nCr = n!/r!*(n-r)!
For more on this, check out Combinatorics & Probability by Veritas Prep)
2. There are 5 truffles. You need to select 2 and not select the rest of the three. Y - select a truffle. N - Do not select it
So you can select 2 truffles in the number of ways in which you can arrange YYNNN.
YYNNN means you select the first 2 and leave the rest.
YNYNN means you select 1st and 3rd and leave the rest.
etc...
This can be done in 5!/2!*3!
You can arrange 5 things in 5! ways. But you have 2 Ys and 3 Ns so you divide by 2! and 3!.
(Again, for more on this, check out Combinatorics & Probability by Veritas Prep)
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Re: MGMAT Word Translations COMBINATORICS!!!  [#permalink]

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New post 21 Jun 2011, 04:37
Yeah I was doing so well in Quant up until I hit Guide 4 of MGMAT Combinatorics...they did an adaquete job at explaining it so I bought the Veritas Combinatorics book from amazon last night! I will put this to rest for a few days I guess and move on to statistics.

Thanks and I kind of understand the nCr thing but I want to go more in depth of the concept so I have a strong foundation.

Thanks!
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Re: MGMAT Word Translations COMBINATORICS!!!  [#permalink]

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New post 21 Jun 2011, 13:34
Project GMAT - also Veritas covers these fairly well, also
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Re: Amy and Adam are making boxes of truffles to give out as wed  [#permalink]

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New post 20 Jul 2013, 10:49
Hi, I have an issue with comprehending such problems. Could any of you correct me please. I know the solution to the problem is 10.

My approach is : Each box can contain two truffles, so if I say there are two slots in the box then the first slot has the possibility of 5 different types of truffles and the next slot has the possibility of 4 different types of truffles. So i concluded 5*4 = 20 as the answer.

Can any of you please identify the flaw in my approach to the problem. Again, I know the solution, but I want to know what exactly is the flaw in the argument presented above, so that I don't repeat the mistake again.
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Re: Amy and Adam are making boxes of truffles to give out as wed  [#permalink]

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New post 20 Jul 2013, 11:14
1
kiranck007 wrote:
Hi, I have an issue with comprehending such problems. Could any of you correct me please. I know the solution to the problem is 10.

My approach is : Each box can contain two truffles, so if I say there are two slots in the box then the first slot has the possibility of 5 different types of truffles and the next slot has the possibility of 4 different types of truffles. So i concluded 5*4 = 20 as the answer.

Can any of you please identify the flaw in my approach to the problem. Again, I know the solution, but I want to know what exactly is the flaw in the argument presented above, so that I don't repeat the mistake again.


The problem with your solution is that 20 has duplications in it: box holding {A, B} is the same as the box holding {B, A} and 5*4 will give you both boxes.

The question basically asks: in how many ways we can select 2 different truffles out of 5, which is simply \(C^2_5=10\).

Hope it's clear.
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Re: Amy and Adam are making boxes of truffles to give out as wed  [#permalink]

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New post 20 Jul 2013, 11:17
1
kiranck007 wrote:
Hi, I have an issue with comprehending such problems. Could any of you correct me please. I know the solution to the problem is 10.

My approach is : Each box can contain two truffles, so if I say there are two slots in the box then the first slot has the possibility of 5 different types of truffles and the next slot has the possibility of 4 different types of truffles. So i concluded 5*4 = 20 as the answer.

Can any of you please identify the flaw in my approach to the problem. Again, I know the solution, but I want to know what exactly is the flaw in the argument presented above, so that I don't repeat the mistake again.


I think you are using the slot-method. I am not very familiar with it, but you are missing the last step:

_ _ , two slots that can be filled in 5*4=20 ways. Correct till here, but then you have to divide for the "factorial" number of interchangeable slots, in this case \(2!=2\). So the final result is \(\frac{5*4}{2}=10\).
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Re: Amy and Adam are making boxes of truffles to give out as wed  [#permalink]

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New post 09 Jul 2014, 14:21
Can someone explain the differences in the approaches you would take if you could pick 2 of the same type of truffle vs how in this problem you can only use one type of each truffle per box?

I understand with the anagram method the latter is YYNNN - out of five choices, two of those five will be picked and three of those 5 will not be picked, therefore the answer is 5!/2!3!

But what about if dupes were allowed, how would the calculation change? Obviously that's a pretty easy thing to manually add but I'd like to understand the process.
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Amy and Adam are making boxes of truffles to give out as wed  [#permalink]

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New post 30 Jun 2018, 02:14
VeritasPrepKarishma niks18 pushpitkc pikolo2510 chetan2u

What is incorrect in below approach:
Category 1 of truffle can be selected in 5C1 (out of 5 different varieties, I need to choose 1)
Category 2 of truffle can be selected in 4C1 (out of 4 available varieties, I need to choose 1)

Hence: Final ans: 5C1 * 4C1
AND means to multiply since I want two different types of truffle in box
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Re: Amy and Adam are making boxes of truffles to give out as wed  [#permalink]

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New post 30 Jun 2018, 02:20
1
adkikani wrote:
VeritasPrepKarishma niks18 pushpitkc pikolo2510 chetan2u

What is incorrect in below approach:
Category 1 of truffle can be selected in 5C1 (out of 5 different varieties, I need to choose 1)
Category 2 of truffle can be selected in 4C1 (out of 4 available varieties, I need to choose 1)

Hence: Final ans: 5C1 * 4C1
AND means to multiply since I want two different types of truffle in box



Hi...

By doing 5C1*4C1, you are choosing two -- one after another
Say five different types are A,B,C,D,E
But here the order does not matter AB in the box is same as BA in the box
So either divide your result by 2 so 5*4/2=10
Or choose 2 out of 5 = 5C2=5!/(3!2!)=10
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Re: Amy and Adam are making boxes of truffles to give out as wed  [#permalink]

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New post 30 Jun 2018, 02:22
1
adkikani wrote:
VeritasPrepKarishma niks18 pushpitkc pikolo2510 chetan2u

What is incorrect in below approach:
Category 1 of truffle can be selected in 5C1 (out of 5 different varieties, I need to choose 1)
Category 2 of truffle can be selected in 4C1 (out of 4 available varieties, I need to choose 1)

Hence: Final ans: 5C1 * 4C1
AND means to multiply since I want two different types of truffle in box


Hi adkikani

Lets say the five truffles are A,B,C,D, and E

Category one - A | Category two - C
is the same as
Category one - C | Category two - A

So, the final possibilities for the truffles in the box are \(\frac{C_1^5 * C_1^4}{2} = 5*\frac{4}{2} = 10\)

Hope this clears your confusion
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Re: Amy and Adam are making boxes of truffles to give out as wed  [#permalink]

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New post 30 Jun 2018, 02:29
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adkikani wrote:
VeritasPrepKarishma niks18 pushpitkc pikolo2510 chetan2u

What is incorrect in below approach:
Category 1 of truffle can be selected in 5C1 (out of 5 different varieties, I need to choose 1)
Category 2 of truffle can be selected in 4C1 (out of 4 available varieties, I need to choose 1)

Hence: Final ans: 5C1 * 4C1
AND means to multiply since I want two different types of truffle in box


By using this method, AB and BA are two different combinations. But according to the question, position doesn't matter. Hence you need to divide by 2 to remove the duplicates

The other way to solve this , just use combination formula 5C2. Answer is 10
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Re: Amy and Adam are making boxes of truffles to give out as wed &nbs [#permalink] 30 Jun 2018, 02:29
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