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Combinatorics example question (Two methods..different results)

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Joined: 24 Apr 2017
Posts: 13
Combinatorics example question (Two methods..different results)  [#permalink]

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25 Jan 2019, 00:35
Hi Team,
I'm reviewing combinators questions. I came to a question that I solved my way and the book solved it another way.

The question is (I changed the names and items a bit):
John is maxing boxes of grapes to give to his friends. He has unlimited supply of 5 different grape colors. If each box has 2 grapes of different colors, how many boxes can he make?

First grape has 5 options. Second grade has 4 options. In total 5*4 = 20.

Another solution (correct one) for the problem is by anagram YYNNN = 5!/(2!3!)=10

Question 1: So my question is when to answer using the first method and when by the second method?
Question 2: The two solutions differ by a factor of 2. If numbers of grapes in the box change from 2 to 3, the solution will differ by a factor of 6! What's the catch?
Thank you
Math Expert
Joined: 02 Aug 2009
Posts: 7334
Re: Combinatorics example question (Two methods..different results)  [#permalink]

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25 Jan 2019, 18:22
mekhdi wrote:
Hi Team,
I'm reviewing combinators questions. I came to a question that I solved my way and the book solved it another way.

The question is (I changed the names and items a bit):
John is maxing boxes of grapes to give to his friends. He has unlimited supply of 5 different grape colors. If each box has 2 grapes of different colors, how many boxes can he make?

First grape has 5 options. Second grade has 4 options. In total 5*4 = 20.

Another solution (correct one) for the problem is by anagram YYNNN = 5!/(2!3!)=10

Question 1: So my question is when to answer using the first method and when by the second method?
Question 2: The two solutions differ by a factor of 2. If numbers of grapes in the box change from 2 to 3, the solution will differ by a factor of 6! What's the catch?
Thank you

Hi..

The two methods you talk of..
(1) 5*4 is nothing but 5P2 or 5C2*2!. So this method talks of ways where order matters and us a Permutation problem. Will work where you have to pick two person out of 5 for post of president and vice President.
But not here, because this is selection of two types of grapes and not arrangements of 2 types of grapes.
So this is wrong.
(2) 5/3!2! Is a combination problem that is ways where order does not matter and is correct here.

Hope it helps clarifying your query
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html
4) Base while finding % increase and % decrease : https://gmatclub.com/forum/percentage-increase-decrease-what-should-be-the-denominator-287528.html

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Manhattan Prep Instructor
Joined: 04 Dec 2015
Posts: 689
GMAT 1: 790 Q51 V49
GRE 1: Q170 V170
Re: Combinatorics example question (Two methods..different results)  [#permalink]

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26 Jan 2019, 13:33
mekhdi wrote:
Hi Team,
I'm reviewing combinators questions. I came to a question that I solved my way and the book solved it another way.

The question is (I changed the names and items a bit):
John is maxing boxes of grapes to give to his friends. He has unlimited supply of 5 different grape colors. If each box has 2 grapes of different colors, how many boxes can he make?

First grape has 5 options. Second grade has 4 options. In total 5*4 = 20.

Another solution (correct one) for the problem is by anagram YYNNN = 5!/(2!3!)=10

Question 1: So my question is when to answer using the first method and when by the second method?
Question 2: The two solutions differ by a factor of 2. If numbers of grapes in the box change from 2 to 3, the solution will differ by a factor of 6! What's the catch?
Thank you

When you solved the problem the first way, you actually counted all of the boxes twice. To see why, try calling the grapes A, B, C, D, and E. Here are the '5 times 4' options:

A (B, C, D, E)
B (A, C, D, E)
C (A, B, D, E)
D (A, B, C, E)
E (A, B, C, D)

For instance, the first line represents the options AB, AC, AD, AE; the second line represents the options BA, BC, BD, BE; etc.

But, look at that again. We counted AB, and we also counted BA separately. We counted AC (on the first line), but we also counted CA. And so on. You don't actually want to count those options twice, because those aren't actually unique boxes: a box with A and B is the same as a box with B and A.

That's why you need to divide your answer by 2 when you approach it this way: to compensate for the fact that you counted every option twice, instead of just once.
_________________

Chelsey Cooley | Manhattan Prep | Seattle and Online

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Re: Combinatorics example question (Two methods..different results)   [#permalink] 26 Jan 2019, 13:33
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