How Do You Know When Order Matters With Dice

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Re: How many dissimilar combinations of outcomes can you make by [#permalink]   18 May 2014, 21:10

NGGMAT wrote:

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

Allow me add my thought too here: When order matters, substantially you are saying that the dice are distinct, say of three different colors. So a 2 on the red dice and 1 on the others is different from ii on the yellowish one with 1 on the other ii.
When order doesn't thing, information technology implies that the dice are identical. If y'all have three identical dice and y'all throw them, a 2, 1, 1 is the aforementioned no thing 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 past rolling [#permalink]   03 Sep 2014, 08:39

eshan429 wrote:

How many dissimilar combinations of outcomes tin can you make past rolling iii standard (6-sided) die if the order of the die does not matter?

(A) 24
(B) thirty
(C) 56
(D) 120
(E) 216

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

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

3) 3 dice are all different:
you have \(half dozen*5*4/iii! = 20\).
Because the question says the order does non affair, you have to split up it by 3!.

and so totally you accept 56.

Hence C.
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Re: How many different combinations of outcomes can you make past [#permalink]   21 Oct 2014, eighteen:00

nitinneha wrote:

Could someone explicate please what is meant by" order doesn't matter" here?

Then that ways that having a different order, doesn't make it a different combination. For example, 1,2,3 OR 3,2,i OR 2,1,iii all just count equally one combination
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Re: How many different combinations of outcomes can you make past [#permalink]   13 Oct 2016, 20:03

Idea most this for a while and this is what I understand by "order doesn't matter"

i. At that place are half dozen*v*four = 120 combinations possible.

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

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

Hope this helps!

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Re: How many dissimilar combinations of outcomes can you make by [#permalink]   01 Nov 2017, 00:19

Bunuel wrote:

rohitgoel15 wrote:

mm007 wrote:

How many different combinations of outcomes can y'all make by rolling three standard (six-sided) dice if the order of the dice does not matter?

(A) 24
(B) 30
(C) 56
(D) 120
(E) 216

The simply style I can see this is 6*half-dozen*6 - six (same outcomes) ... Tin can anyone explain?

If the social club of the dice does not matter then we tin can accept three cases:
1. 30 - all die show akin numbers: half-dozen outcomes (111, 222, ..., 666);
2. XXY - two dice show akin numbers and third is different: \(half dozen*five=30\), half dozen choices for X and v choices for Y;
iii. XYZ - all three dice show distinct numbers: \(C^3_6=twenty\), selecting three unlike numbers from 6;

Total: half-dozen+thirty+20=56.

Answer: C.

How-do-you-do
Tin can I get any link which make me piece of cake to sympathise this topic boz I one thousand very much confused.
Kindly help me combination and pick are very confusing topics.
Thx

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Re: How many different combinations of outcomes can you brand past [#permalink]   01 November 2017, 00:25

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Re: How many dissimilar combinations of outcomes can you make by [#permalink]   02 Nov 2017, 15:59

mm007 wrote:

How many different combinations of outcomes can you make past rolling iii standard (six-sided) dice if the lodge of the dice does non matter?

(A) 24
(B) 30
(C) 56
(D) 120
(E) 216

When nosotros coil 3 dice, we could have the following 3 cases:

1) iii of the aforementioned numbers

two) 2 of the same numbers and 1 unlike number

3) all 3 different numbers

Let's analyze each of these cases.

i) 3 of the same numbers

If all 3 numbers are the aforementioned, they can merely be (ane, ane, 1), (2, 2, 2), ... , (6, 6, 6). Thus, there are half dozen outcomes in this case.

2) 2 of the same numbers and 1 different number

If 2 numbers are the aforementioned and the third one is different, they could exist, for case, two 1s and one of the five other numbers. However, we can replace the two 1s with any of the 5 numbers. In other words, we accept six choices for the 2 numbers that are the same and 5 choices for the number that is different. Thus, there are six x v = thirty outcomes in this example.

3) all 3 different numbers

Since we are picking 3 numbers from six in which club doesn't matter, there are 6C3 = 6!/[3!(vi-3)!] = (6 x 5 x 4)/3! = (6 ten 5 x iv)/half-dozen = 20 outcomes in this case.

Therefore, there are vi + thirty + 20 = 56 outcomes in total.

Answer: C
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How many dissimilar combinations of outcomes can yous brand by [#permalink]   12 Nov 2017, 22:22

mm007 wrote:

How many different combinations of outcomes can you make by rolling 3 standard (6-sided) die if the order of the die does not affair?

(A) 24
(B) 30
(C) 56
(D) 120
(Eastward) 216

When social club is important we have say, 112 unlike from 121 or 211. If gild is not of import all three are counted as ane upshot.

At present, all the three digits may be different, ii digits may be the same and i different or all the three aforementioned.

In the first instance, there are 6C3 = 20 combinations, in the second case 6*5=30 combinations and in the 3rd example half dozen combinations for a total of 56 combinations.
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Re: How many different combinations of outcomes can yous make by [#permalink]   23 Apr 2018, 19:42

VeritasPrepKarishma wrote:

NGGMAT wrote:

and when would information technology be 6*6*6??

Let me add together my thought too here: When social club matters, essentially yous are saying that the dice are singled-out, say of iii different colors. Then 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 y'all throw them, a 2, 1, ane is the same no affair which die gives you 2 because you cannot tell them apart.

But why do we non carve up 6*five by 2?
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Re: How many different combinations of outcomes can y'all brand by [#permalink]   23 April 2018, twenty:24

dannythor6911 wrote:

VeritasPrepKarishma wrote:

NGGMAT wrote:

and when would it be 6*half-dozen*6??

Let me add my thought too here: When gild matters, essentially you lot are saying that the die are singled-out, say of three different colors. So a 2 on the red dice and one 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 3 identical dice and you throw them, a ii, ane, 1 is the same no matter which die gives y'all 2 because you cannot tell them apart.

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

Because information technology is given that the gild DOES NOT matter.

When we adjust XYY in 3! ways and divide by two, it is considering when counting iii!, we take assumed that the 2 Ys are singled-out so XY1Y2 and XY2Y1 are two dissimilar arrangements (Y1 is 2nd position and Y2 on tertiary as against Y2 on second position and Y1 on third). Only actually there is only one arrangement XYY. So the number of arrangements becomes half.
Here there is no arrangement since the 3 places __ __ __ are non singled-out. Think that the three 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 tin you lot brand by [#permalink]   04 May 2018, 20:25

Bunuel wrote:

rohitgoel15 wrote:

mm007 wrote:

How many different combinations of outcomes can you make past rolling three standard (vi-sided) dice if the order of the dice does not thing?

(A) 24
(B) 30
(C) 56
(D) 120
(East) 216

The simply way I can see this is vi*6*6 - 6 (aforementioned outcomes) ... Can anyone explain?

If the guild of the dice does non thing then we can have 3 cases:
1. Thirty - all dice show alike numbers: 6 outcomes (111, 222, ..., 666);
ii. XXY - two dice testify alike numbers and third is different: \(6*five=30\), 6 choices for Ten and 5 choices for Y;
3. XYZ - all three dice show distinct numbers: \(C^3_6=twenty\), selecting three dissimilar numbers from 6;

Total: 6+30+twenty=56.

Answer: C.

Is there some way to solve from the total if the order matters?

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Re: How many different combinations of outcomes can yous make by [#permalink]   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.east. the dice are distinct, there are 6 possible outcomes for each dice.
The total number of possible outcomes is so simply half dozen*half-dozen*6 = 216
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Re: P and C [#permalink]   24 Oct 2018, 06:45

uttam94317 wrote:

How many different combinations of outcomes can you brand past rolling three standard (6-sided) dice if the
social club of the dice does not matter?
(A) 24 (B) 30 (C) 56 (D) 120 (East) 216

If the gild of the dice don't matter and then we tin distinguish the outcomes in post-obit ways

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

Case 2: When two dice are same and third is different - 6C2*2 = 15*two = 30 ways (Cull 2 out of 6 possible issue of dice in 6C2 means, then choose the number out of two outcomes that repeats due east.grand. 225 or 255 way if 2 called numbers are 2 and v)

Instance iii: When all the die show same effect - six ways (111 or 222 or 333 etc.)

Total favourable cases = 20+30+six = 56 means

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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Re: How many different combinations of outcomes tin can you make past [#permalink]   25 Oct 2018, 16:54

mm007 wrote:

How many different combinations of outcomes can you lot make by rolling three standard (6-sided) dice if the social club of the dice does not matter?

(A) 24
(B) 30
(C) 56
(D) 120
(E) 216

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

If two of the three numbers are the same and the 3rd is unlike, we take 6C2 x 2 = (6 x five)/2 x two = 30 combinations. (Annotation: There are 6C2 ways to cull ii numbers from 6 when guild doesn't matter. However, when two numbers are chosen, for case, ane and ii, the combinations (1, ane, 2) and (2, two, one) are considered dissimilar, therefore, we need to multiply by ii.)

If all 3 numbers are the unlike, we have 6C3 = (half dozen x 5 x 4)/(iii 10 2) = 20 combinations.

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

Answer: C
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Re: How many different combinations of outcomes tin can you make past [#permalink]   23 Jul 2020, 22:07

Bunuel wrote:

rohitgoel15 wrote:

mm007 wrote:

How many different combinations of outcomes can you brand past rolling iii standard (6-sided) die if the order of the die does non matter?

(A) 24
(B) xxx
(C) 56
(D) 120
(Eastward) 216

The only mode I tin can run into this is 6*6*vi - 6 (same outcomes) ... Can anyone explain?

If the order of the dice does not matter so we can have 3 cases:
1. Xxx - all dice bear witness alike numbers: vi outcomes (111, 222, ..., 666);
2. XXY - ii die show akin numbers and tertiary is different: \(6*5=30\), six choices for X and 5 choices for Y;
3. XYZ - all iii dice show distinct numbers: \(C^3_6=twenty\), selecting three different numbers from 6;

Total: 6+30+20=56.

Answer: C.

The case 2 of your solution seems wrong because half-dozen*5=30 and in these 30 sequences lodge matters. And then to correct that we accept to divide it by 2 and the number of combination from case 2 should be xv.
Kindly, permit me know what i am missing here.

Thanks

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Re: How many dissimilar combinations of outcomes can you make by [#permalink]   23 Jul 2020, 23:48

manikmehta95 wrote:

Bunuel wrote:

rohitgoel15 wrote:

mm007 wrote:

How many dissimilar combinations of outcomes can you lot make past rolling three standard (6-sided) dice if the order of the dice does not matter?

(A) 24
(B) 30
(C) 56
(D) 120
(East) 216

The only way I tin meet this is 6*6*half dozen - 6 (same outcomes) ... Can anyone explicate?

If the order of the dice does not matter then we can have three cases:
1. Xxx - all dice show alike numbers: 6 outcomes (111, 222, ..., 666);
ii. XXY - 2 dice evidence akin numbers and 3rd is dissimilar: \(6*5=thirty\), half dozen choices for Ten and 5 choices for Y;
iii. XYZ - all three dice show singled-out numbers: \(C^3_6=20\), selecting three different numbers from 6;

Full: six+30+twenty=56.

Answer: C.

The case two of your solution seems wrong because half dozen*5=30 and in these xxx sequences lodge 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 i 2
1 one three
ane ane 4
1 one 5
one ane half-dozen
two 2 1
two ii 3
2 2 4
2 2 v
2 2 vi
3 iii one
3 3 2
3 3 4
iii 3 5
3 3 half-dozen
iv 4 1
iv 4 2
four 4 3
four 4 5
4 4 6
5 v 1
5 5 2
five v 3
5 5 four
5 five 6
half dozen 6 i
6 6 2
6 half dozen 3
6 6 4
6 half dozen five

AS you tin can see no duplication there.
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How many different combinations of outcomes tin can yous brand by [#permalink]   24 Jul 2020, 08:33

Solution

This question can be solved in dissimilar means.
I idea of sharing an approach that can relieve a lot of time.

Given

• iii standard 6-sided dice are rolled.

To Detect

• No of combinations when order does non matter.

Approach and Working Out

• Let'southward 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 = eastward
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-one)
o = 8C5
o = 56

• Delight 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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Re: How many different combinations of outcomes can you brand by [#permalink]   18 Jun 2021, 23:33

Why is the following incorrect?

1. Total number of possible unique outcomes: 6*half-dozen*half dozen = 216 , i.e. (whatever number of the three dice may come up up) - Primal Rule

Since nosotros accept all possible unique outcomes, we just need to focus on finding the number of sets.

A fix will have half dozen unique outcomes : {1,two,3} | {1,3,2} | {2,1,three} | {2,3,1} | {three,1,2} | {3,two,one}

Since nosotros know that one set has 6 unique outcomes, and we have a full of 216 unique outcomes. The number of possible sets = 216/half dozen = 36 sets

Kindly assist

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Re: How many different combinations of outcomes can you brand by [#permalink]   25 Jun 2021, 07:55

tennis1ball wrote:

C. 56.

1) All dice accept the aforementioned number:
y'all have vi possibilities.

2) 2 dice have the same number, merely the 3rd is different:
you take 6*5

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

then totally yous have 56.

in instance 2 also order doesn't matter.but why we did not separate by iii?

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Re: How many unlike combinations of outcomes can y'all make past [#permalink]   27 Jun 2021, 02:27

Bunuel wrote:

rohitgoel15 wrote:

mm007 wrote:

How many dissimilar combinations of outcomes tin can you make past rolling three standard (6-sided) dice if the guild of the die does not affair?

(A) 24
(B) 30
(C) 56
(D) 120
(Eastward) 216

The only manner I can encounter this is 6*6*6 - vi (same outcomes) ... Can anyone explain?

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

Total: half-dozen+thirty+20=56.

Answer: C.

Hullo Bunuel

Cheers for the amazing explanation, Simply for the sake of conceptual clarity could you too show us a bridge between 216 (when we accept 6x6x6) and 56 (answer) equally to how to we arrive at 56 by excluding the things that nosotros don't have to count from 216 such every bit Unique repetitions, non unique repetitions etc. Would exist actually helpful for me to get conceptual clarity

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

27 Jun 2021, 02:27

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