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# rolling 3 standard dices

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Manager
Status: Still Struggling
Joined: 02 Nov 2010
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Location: India
GMAT Date: 10-15-2011
GPA: 3.71
WE: Information Technology (Computer Software)
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Kudos [?]: 6 [0], given: 8

rolling 3 standard dices [#permalink]  04 Nov 2010, 10:51
00:00

Difficulty:

5% (low)

Question Stats:

27% (01:36) correct 72% (00:39) wrong based on 11 sessions
How many different combinations of outcomes can you make by rolling three standard (6-sided) dice if the order of the dice does not matter?
(a) - 24
(b) - 30
(c) - 56
(d) - 120
(e) - 216
[Reveal] Spoiler: OA

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Manager
Joined: 10 Sep 2010
Posts: 133
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Kudos [?]: 10 [0], given: 7

Re: rolling 3 standard dices [#permalink]  04 Nov 2010, 13:17
First I though that it is easy: 6x6x6=216. Then I checked the answer and it was wrong.
I started to think.

Here it goes:

The number of combinations will include: xxx, xyy (which is the same as xxy), xyz

xxx - all cubes showing the same number - 6 possible combinations
xyy - one cube shows one number, while two other cubes show the same number, that is different from what the first cube shows - for each cube showing number there are 5 other numbers left that can be shown by two other cubes - 5x6
xxy - two cubes show the same number, while the third cube shows different number (which is the same as previous)
xyz - all cubes show different numbers - 6x5x4/3x2=4x5.

6+(6x5)+(4x5)=56
Manager
Joined: 10 Sep 2010
Posts: 133
Followers: 2

Kudos [?]: 10 [0], given: 7

Re: rolling 3 standard dices [#permalink]  04 Nov 2010, 13:26
It felt like this problem was harder than 600-700 level...
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Re: rolling 3 standard dices [#permalink]  04 Nov 2010, 14:08
1
KUDOS
Three dice show three diff numbers - C(6,3) - choose 3 numbers ... 20 ways

Similarly 2 same 1 diff number - C(6,2) - 30 ways

Same number - C(6,1) - 6 ways

Total - 56

Posted from my mobile device
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Math write-ups
1) Algebra-101 2) Sequences 3) Set combinatorics 4) 3-D geometry

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Manager
Joined: 10 Sep 2010
Posts: 133
Followers: 2

Kudos [?]: 10 [0], given: 7

Re: rolling 3 standard dices [#permalink]  04 Nov 2010, 14:24
shrouded1 wrote:
Three dice show three diff numbers - C(6,3) - choose 3 numbers ... 20 ways

Similarly 2 same 1 diff number - C(6,2) - 30 ways

Same number - C(6,1) - 6 ways

Total - 56

Posted from my mobile device

This is much better explanation, than mine. Really straightforward.
Great. Thanks.
Manager
Status: Still Struggling
Joined: 02 Nov 2010
Posts: 139
Location: India
GMAT Date: 10-15-2011
GPA: 3.71
WE: Information Technology (Computer Software)
Followers: 4

Kudos [?]: 6 [0], given: 8

Re: rolling 3 standard dices [#permalink]  06 Nov 2010, 02:19
shrouded1 wrote:
Three dice show three diff numbers - C(6,3) - choose 3 numbers ... 20 ways

Similarly 2 same 1 diff number - C(6,2) - 30 ways

Same number - C(6,1) - 6 ways

Total - 56

can u pls be a little more elaborate as on why u r getting C(6,3) for 2 dices show diff numbers?
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Re: rolling 3 standard dices [#permalink]  06 Nov 2010, 03:08
Expert's post
krishnasty wrote:
How many different combinations of outcomes can you make by rolling three standard (6-sided) dice if the order of the dice does not matter?
(a) - 24
(b) - 30
(c) - 56
(d) - 120
(e) - 216

First, if the dice were all unique, we could have calculated the number of combinations using basic counting principles i.e. we could have said that dice 1 can show 1/2/3/4/5/6 so 6 different outcomes, dice 2 has 6 different outcomes and dice 3 has 6 different outcomes. So possible outcomes = 6*6*6 = 216.

But we cannot do it here because the dice are identical. Also, I cannot say that since they are identical, let me just divide 216 by 3!. The reason is that not every possible outcome is counted 3! times when the dice are unique e.g.

_1_ _2_ _4_
D1 D2 D3

_1_ _4_ _2_
D1 D2 D3
and so on... Using 1, 2 and 4, we have 6 possible combinations.
But using 1, 2 and 2 we have only 3 possible combinations when the dice are unique and using 2, 2 and 2 there is only one possible combination. When we divide the entire 216 by 6!, we are reducing each of these combinations by a factor of 6 which is incorrect and that is why you get 216/6 = 36 i.e. less combinations than there actually are.

Then what to do in such questions?
Method 1: The better method is the one suggested above.
The solution of the dice can take three different forms.
Form 1: Each number is distinct e.g. 1, 2 and 4 above.
Out of 6 numbers (1/2/3/4/5/6), you select 3 using 6C3 = 20. You do not have to arrange these numbers since the dice are identical. So you will now get all such cases: (1,2, 3), (1, 2, 4), (1, 2, 5) ....(3, 4, 5)....(4, 5, 6)
Form 2: Two numbers are identical, one is different.
Out of 6 numbers (1/2/3/4/5/6), you select 1 which is going to be the identical one and then select another from the rest 5 numbers which will be the different one. So you get 6C1 * 5C1 = 30 combinations. Now you get all such cases: (1, 1, 2), (1, 1, 3), ....(2, 2, 1) ... (6, 6, 5)

Form 3: All three numbers are identical. Since there are 6 numbers, you can choose any one number in 6 ways. Now you will get all such cases: (1, 1, 1), (2, 2, 2)....(6, 6, 6)

Total possible combinations: 20 + 30 + 6 = 56

Worst case scenario: Let's say, it doesn't strike me at the time of the exam. The numbers are not so unwieldy that you cannot enumerate.
Just keep 'pattern recognition' in mind.
I say, I need to count all possible combinations.
(1, 1, 1), (1, 1, 2), (1, 1, 3) ... = 6
(1, 2, 2), (1, 2, 3)... = 5 (Here I do not have to count (1, 2, 1) because it is already taken above)
(1, 3, 3), (1, 3, 4) ... = 4
I see the pattern here. Numbers of the form (1, x, y) will be 6 + 5 + 4 + 3 + 2 + 1 = 21

Now (2, 2, 2), (2, 2, 3).... = 5 (I begin with 2, 2, 2 because I do not have to take a number lower than the one on the left i.e. I cannot take (2, 2, 1) since it has already been counted above as (1, 2, 2))
(2, 3, 3), (2, 3, 4) .... = 4
Again I see the pattern. Numbers of the form (2, x, y) will be 5 + 4 + 3 + 2 + 1 = 15
Which means numbers of the form (3, x, y) will be 4 + 3 + 2 + 1 = 10
Of the form (4, x, y) will be = 3 + 2 + 1 = 6
Of the form (5, x , y) will be = 2 + 1 = 3
Of the form (6, x, y) will be = 1

Here x is greater than or equal to the leftmost number and y is greater than or equal to x.

A painful method but do-able in 2 minutes if you recognize the pattern. To be used in worst case scenario.
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Karishma
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Save $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Senior Manager Joined: 08 Nov 2010 Posts: 424 WE 1: Business Development Followers: 6 Kudos [?]: 28 [0], given: 161 Re: rolling 3 standard dices [#permalink] 08 Nov 2010, 06:17 Hey guys. Thanks, but i have a stupid question... when u say 6c3 - what do u mean? 6c3 = 20 - what is the calculation? i read the explanation for this question 15 times - and no luck... help me plz... Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 4028 Location: Pune, India Followers: 858 Kudos [?]: 3613 [1] , given: 144 Re: rolling 3 standard dices [#permalink] 08 Nov 2010, 10:36 1 This post received KUDOS Expert's post 144144 wrote: Hey guys. Thanks, but i have a stupid question... when u say 6c3 - what do u mean? 6c3 = 20 - what is the calculation? i read the explanation for this question 15 times - and no luck... help me plz... nCr = \frac{n!}{(r!*(n-r)!)} So 6C3 = \frac{6!}{(3!*3!)} We use this formula when out of n distinct things, you have to select r. My suggestion would be to first go through the complete theory of Permutations and Combinations. Then it will be much easier to understand the explanations. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting
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Senior Manager
Joined: 08 Nov 2010
Posts: 424
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Kudos [?]: 28 [0], given: 161

Re: rolling 3 standard dices [#permalink]  08 Nov 2010, 12:09

I like the way you answer the questions and i'm personally very gratful

thanks.
Veritas Prep GMAT Instructor
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Re: rolling 3 standard dices [#permalink]  09 Nov 2010, 09:35
Expert's post
Thanks. We will do well if we use the 'Big Picture Approach' for problems. That is, once we are done with a particular question, we should be able to answer not only any similar question, but also any question with a slight variation in the concept. So it is important to understand the 'why' of every step and important to think 'what if I changed this to this'...
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Re: rolling 3 standard dices   [#permalink] 09 Nov 2010, 09:35
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