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How many different combinations of outcomes can you make by

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Re: How many different combinations of outcomes can you make by  [#permalink]

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New post 18 May 2014, 21:10
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NGGMAT wrote:

and when would it be 6*6*6??


Let me add my thought too here: When order matters, essentially you are saying that the dice are distinct, say of three different colors. So a 2 on the red die and 1 on the others is different from 2 on the yellow one with 1 on the other two.
When order doesn't matter, it implies that the dice are identical. If you have three identical dice and you throw them, a 2, 1, 1 is the same no matter which die gives you 2 because you cannot tell them apart.
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Re: How many different combinations of outcomes can you make by rolling  [#permalink]

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New post 03 Sep 2014, 08:39
1
eshan429 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

1) All dice have the same number:
you have \(6\) possibilities.

2) 2 dice have the same number, but the 3rd is different:
you have \(6*5 = 30\)

3) 3 dice are all different:
you have \(6*5*4/3! = 20\).
Because the question says the order does not matter, you have to divide it by 3!.

so totally you have 56.

Hence C.
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New post 21 Oct 2014, 18:00
nitinneha wrote:
Could someone explain please what is meant by" order doesn't matter" here?


So that means that having a different order, doesn't make it a different combination. For example, 1,2,3 OR 3,2,1 OR 2,1,3 all only count as 1 combination
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Re: How many different combinations of outcomes can you make by  [#permalink]

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New post 13 Oct 2016, 20:03
1
Thought about this for a while and this is what I understand by "order doesn't matter"

1. There are 6*5*4 = 120 combinations possible.

2. For a 3 digit number with unique digits if the "Order matters" the 654 645 546 564 456 465 = 6 combinations are possible
Order doesn't matter: You have just one combination.

So for 120 combinations if the order doesn't matter 120/6 = 20.

Hope this helps!
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Re: How many different combinations of outcomes can you make by  [#permalink]

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New post 01 Nov 2017, 00:19
Bunuel wrote:
rohitgoel15 wrote:
mm007 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


The only way I can see this is 6*6*6 - 6 (same outcomes) ... Can anyone explain?


If the order of the dice does not matter then we can have 3 cases:
1. XXX - all dice show alike numbers: 6 outcomes (111, 222, ..., 666);
2. XXY - two dice show alike numbers and third is different: \(6*5=30\), 6 choices for X and 5 choices for Y;
3. XYZ - all three dice show distinct numbers: \(C^3_6=20\), selecting three different numbers from 6;

Total: 6+30+20=56.

Answer: C.

Hi
Can I get any link which make me easy to understand this topic boz I m very much confused.
Kindly help me combination and selection are very confusing topics.
Thx

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New post 01 Nov 2017, 00:25
samnidhi wrote:
Hi
Can I get any link which make me easy to understand this topic boz I m very much confused.
Kindly help me combination and selection are very confusing topics.
Thx

Sent from my A1601 using GMAT Club Forum mobile app


21. Combinatorics/Counting Methods



For more:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread


Hope it helps.
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Re: How many different combinations of outcomes can you make by  [#permalink]

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New post 02 Nov 2017, 15:59
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mm007 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


When we roll 3 dice, we could have the following 3 cases:

1) 3 of the same numbers

2) 2 of the same numbers and 1 different number

3) all 3 different numbers

Let’s analyze each of these cases.

1) 3 of the same numbers

If all 3 numbers are the same, they can only be (1, 1, 1), (2, 2, 2), ... , (6, 6, 6). Thus, there are 6 outcomes in this case.

2) 2 of the same numbers and 1 different number

If 2 numbers are the same and the third one is different, they could be, for example, two 1s and one of the 5 other numbers. However, we can replace the two 1s with any of the 5 numbers. In other words, we have 6 choices for the 2 numbers that are the same and 5 choices for the number that is different. Thus, there are 6 x 5 = 30 outcomes in this case.

3) all 3 different numbers

Since we are picking 3 numbers from 6 in which order doesn’t matter, there are 6C3 = 6!/[3!(6-3)!] = (6 x 5 x 4)/3! = (6 x 5 x 4)/6 = 20 outcomes in this case.

Therefore, there are 6 + 30 + 20 = 56 outcomes in total.

Answer: C
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How many different combinations of outcomes can you make by  [#permalink]

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New post 12 Nov 2017, 22:22
mm007 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

When order is important we have say, 112 different from 121 or 211. If order is not important all three are counted as 1 outcome.

Now, all the three digits may be different, 2 digits may be the same and 1 different or all the three same.

In the first case, there are 6C3 = 20 combinations, in the second case 6*5=30 combinations and in the third case 6 combinations for a total of 56 combinations.
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Re: How many different combinations of outcomes can you make by  [#permalink]

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New post 23 Apr 2018, 19:42
VeritasPrepKarishma wrote:
NGGMAT wrote:

and when would it be 6*6*6??


Let me add my thought too here: When order matters, essentially you are saying that the dice are distinct, say of three different colors. So a 2 on the red die and 1 on the others is different from 2 on the yellow one with 1 on the other two.
When order doesn't matter, it implies that the dice are identical. If you have three identical dice and you throw them, a 2, 1, 1 is the same no matter which die gives you 2 because you cannot tell them apart.


But why do we not divide 6*5 by 2?
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New post 23 Apr 2018, 20:24
dannythor6911 wrote:
VeritasPrepKarishma wrote:
NGGMAT wrote:

and when would it be 6*6*6??


Let me add my thought too here: When order matters, essentially you are saying that the dice are distinct, say of three different colors. So a 2 on the red die and 1 on the others is different from 2 on the yellow one with 1 on the other two.
When order doesn't matter, it implies that the dice are identical. If you have three identical dice and you throw them, a 2, 1, 1 is the same no matter which die gives you 2 because you cannot tell them apart.


But why do we not divide 6*5 by 2?


Because it is given that the order DOES NOT matter.

When we arrange XYY in 3! ways and divide by 2, it is because when counting 3!, we have assumed that the two Ys are distinct so XY1Y2 and XY2Y1 are 2 different arrangements (Y1 is second position and Y2 on third as against Y2 on second position and Y1 on third). But actually there is only one arrangement XYY. So the number of arrangements becomes half.
Here there is no arrangement since the 3 places __ __ __ are not distinct. Think that the 3 dice are completely identical. So you don't have a first-second-third position.
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Re: How many different combinations of outcomes can you make by  [#permalink]

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New post 04 May 2018, 20:25
Bunuel wrote:
rohitgoel15 wrote:
mm007 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


The only way I can see this is 6*6*6 - 6 (same outcomes) ... Can anyone explain?


If the order of the dice does not matter then we can have 3 cases:
1. XXX - all dice show alike numbers: 6 outcomes (111, 222, ..., 666);
2. XXY - two dice show alike numbers and third is different: \(6*5=30\), 6 choices for X and 5 choices for Y;
3. XYZ - all three dice show distinct numbers: \(C^3_6=20\), selecting three different numbers from 6;

Total: 6+30+20=56.

Answer: C.


Is there some way to solve from the total if the order matters?
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New post 04 May 2018, 21:44
Mco100 wrote:
Is there some way to solve from the total if the order matters?


If the order matters i.e. the dice are distinct, there are 6 possible outcomes for each die.
The total number of possible outcomes is then simply 6*6*6 = 216
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New post 24 Oct 2018, 06:45
uttam94317 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


If the order of the dice don't matter then we can distinguish the outcomes in following ways

Case 1: When all the three dice show Different outcomes on them - This may happen in 6C3 ways = 20 ways (Choosing 3 distinct outcomes out of 6 possible outcomes on each dice)

Case 2: When two dice are same and third is different - 6C2*2 = 15*2 = 30 ways (Choose 2 out of 6 possible outcome of dice in 6C2 ways, then choose the number out of two outcomes that repeats e.g. 225 or 255 way if two chosen numbers are 2 and 5)

Case 3: When all the dice show same outcome - 6 ways (111 or 222 or 333 etc.)

Total favourable cases = 20+30+6 = 56 ways

Answer: Option C

Bunuel : This question has been discussed here
https://gmatclub.com/forum/how-many-dif ... ml#p280405
so Please merge the topics

uttam94317 Please search the question before you post. Read the rules of posting
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New post 25 Oct 2018, 16:54
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mm007 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


If all 3 numbers are the same, we have 6C1 = 6 combinations.

If two of the 3 numbers are the same and the third is different, we have 6C2 x 2 = (6 x 5)/2 x 2 = 30 combinations. (Note: There are 6C2 ways to choose 2 numbers from 6 when order doesn’t matter. However, when two numbers are chosen, for example, 1 and 2, the combinations (1, 1, 2) and (2, 2, 1) are considered different, therefore, we need to multiply by 2.)

If all 3 numbers are the different, we have 6C3 = (6 x 5 x 4)/(3 x 2) = 20 combinations.

Therefore, there are 6 + 30 + 20 = 56 combinations.

Answer: C
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Re: How many different combinations of outcomes can you make by  [#permalink]

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New post 23 Jul 2020, 22:07
Bunuel wrote:
rohitgoel15 wrote:
mm007 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


The only way I can see this is 6*6*6 - 6 (same outcomes) ... Can anyone explain?


If the order of the dice does not matter then we can have 3 cases:
1. XXX - all dice show alike numbers: 6 outcomes (111, 222, ..., 666);
2. XXY - two dice show alike numbers and third is different: \(6*5=30\), 6 choices for X and 5 choices for Y;
3. XYZ - all three dice show distinct numbers: \(C^3_6=20\), selecting three different numbers from 6;

Total: 6+30+20=56.

Answer: C.


The case 2 of your solution seems wrong because 6*5=30 and in these 30 sequences order matters. So to correct that we have to divide it by 2 and the number of combination from case 2 should be 15.
Kindly, let me know what i am missing here.

Thanks
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New post 23 Jul 2020, 23:48
manikmehta95 wrote:
Bunuel wrote:
rohitgoel15 wrote:
mm007 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


The only way I can see this is 6*6*6 - 6 (same outcomes) ... Can anyone explain?


If the order of the dice does not matter then we can have 3 cases:
1. XXX - all dice show alike numbers: 6 outcomes (111, 222, ..., 666);
2. XXY - two dice show alike numbers and third is different: \(6*5=30\), 6 choices for X and 5 choices for Y;
3. XYZ - all three dice show distinct numbers: \(C^3_6=20\), selecting three different numbers from 6;

Total: 6+30+20=56.

Answer: C.


The case 2 of your solution seems wrong because 6*5=30 and in these 30 sequences order matters. So to correct that we have to divide it by 2 and the number of combination from case 2 should be 15.
Kindly, let me know what i am missing here.

Thanks


Not sure how you deduced that but here are all 30 cases of XXY:
1 1 2
1 1 3
1 1 4
1 1 5
1 1 6
2 2 1
2 2 3
2 2 4
2 2 5
2 2 6
3 3 1
3 3 2
3 3 4
3 3 5
3 3 6
4 4 1
4 4 2
4 4 3
4 4 5
4 4 6
5 5 1
5 5 2
5 5 3
5 5 4
5 5 6
6 6 1
6 6 2
6 6 3
6 6 4
6 6 5

AS you can see no duplication there.
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New post 24 Jul 2020, 08:33

Solution



This question can be solved in different ways.
I thought of sharing an approach that can save a lot of time.

Given

• 3 standard 6-sided dice are rolled.

To Find

• No of combinations when order does not matter.

Approach and Working Out

• Let’s assume the number of 1s, 2s, till 6s.
o No of 1s = a
o No of 2s = b
o No of 3s = c
o No of 4s = d
o No of 5s = e
o No of 6s = f
• We can say, a + b + c + d + e + f = 3 (as we have 3 dices).
• The number of non-negative integral solutions of this equation will be,
o (3 + 6 – 1)C(6-1)
o = 8C5
o = 56

• Please note, the number of non-negative integral solution of an equation a + b + c … + r = n is given by, (n + r – 1)C(r – 1).

Correct Answer: Option C
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