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# How many different combinations of outcomes can you make by

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

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21 Dec 2006, 17:34
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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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Re: PS_How many different ... [#permalink]

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21 Dec 2006, 18:43
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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

IMO Answer is E. 6*6*6 = 216 since order does not matter,

Last edited by 800_gal on 28 Dec 2006, 23:50, edited 1 time in total.
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21 Dec 2006, 19:00
Could someone explain please what is meant by" order doesn't matter" here?

also 3C3=1. So the combinations would be same in either case.

thanks a lot guys!
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21 Dec 2006, 19:45
C 56

I count all the possible combo.

definately not 6*6*6 since the question said 'order does not matters'.
It means we do not distinguish between (1,1,5) or (1,5,1) or (5,1,1) .All these are considered as 1 outcome.
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21 Dec 2006, 21:46
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C. 56.

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

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

so totally you have 56.
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23 Dec 2006, 20:49
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Some further explanation on my answer above:

Case (2) is different from case (3).
In case 2, e.g. 5, (6, 6) and 6, (5, 5) are different.

However, in case (3), e.g. 2, 3, 4, and 4, 2, 3 are the same.

So, in case 2, you don't divide it by 2!, but in case (3), you divide it by 3!.
Case (2) is more of a permutation on itself.
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Re: How many different combinations of outcomes can you make by [#permalink]

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09 Mar 2012, 02:14
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?
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Re: How many different combinations of outcomes can you make by [#permalink]

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09 Mar 2012, 07:26
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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.

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

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08 Sep 2013, 21:36
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.

HI BUnuel,

Could you please explain me this part --

2. XXY - two dice show alike numbers and third is different: $$6*5=30$$, 6 choices for X and 5 choices for Y;

i got 1st and 3rd but this one..

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

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08 Sep 2013, 22:36
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Expert's post
ishdeep18 wrote:
Bunuel wrote:
rohitgoel15 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 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.

HI BUnuel,

Could you please explain me this part --

2. XXY - two dice show alike numbers and third is different: $$6*5=30$$, 6 choices for X and 5 choices for Y;

i got 1st and 3rd but this one..

Regards
Ishdeep Singh

Would be easier to list all possible cases:
XXY
112,
113,
114,
115,
116,

221,
223,
224,
225,
226,

331,
332,
334,
335,
336,

441,
442,
443,
445,
446,

551,
552,
553,
554,
556,

661,
662,
663,
664,
665.

6 choices for X and 5 choices for Y.

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

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30 Nov 2013, 02:41
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.

Brunel , what if order of dice matters ??
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Re: How many different combinations of outcomes can you make by [#permalink]

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17 Apr 2014, 01:54
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.

Hi B,

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;

Could you explain why do we take combination for 3rd choice 3)- $$C^3_6$$ and not for the 2nd choice 2) $$C^2_6$$?

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

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17 Apr 2014, 02:15
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Expert's post
pkhats wrote:
Bunuel wrote:
rohitgoel15 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.

Hi B,

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;

Could you explain why do we take combination for 3rd choice 3)- $$C^3_6$$ and not for the 2nd choice 2) $$C^2_6$$?

Thanks,
PK

For the third case (XYZ) $$C^3_6=20$$ gives all 3-number combinations out of 6, which is exactly what we need for this case:
{1, 2, 3}
{1, 2, 4}
...
{4, 5, 6}

While if we use $$C^2_6=15$$ for the second case (XXY) we get all 2-number combinations out of 6:
{1, 2}
{1, 3}
{1, 4}
...
{5, 6}

But we are rolling 3 dice and we need to take this into account. For example, {1, 2} case can be {1, 1, 2} or {1, 2, 2} and the same for all other cases, which means that we need to multiply $$C^2_6=15$$ by 2 to get the total number of this case.

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

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18 Apr 2014, 01:23
Bunuel wrote:
pkhats wrote:
Bunuel wrote:

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.

Hi B,

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;

Could you explain why do we take combination for 3rd choice 3)- $$C^3_6$$ and not for the 2nd choice 2) $$C^2_6$$?

Thanks,
PK

For the third case (XYZ) $$C^3_6=20$$ gives all 3-number combinations out of 6, which is exactly what we need for this case:
{1, 2, 3}
{1, 2, 4}
...
{4, 5, 6}

While if we use $$C^2_6=15$$ for the second case (XXY) we get all 2-number combinations out of 6:
{1, 2}
{1, 3}
{1, 4}
...
{5, 6}

But we are rolling 3 dice and we need to take this into account. For example, {1, 2} case can be {1, 1, 2} or {1, 2, 2} and the same for all other cases, which means that we need to multiply $$C^2_6=15$$ by 2 to get the total number of this case.

Does this make sense?

Great!! Thanks for the example!
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Re: How many different combinations of outcomes can you make by [#permalink]

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28 Apr 2014, 19:50
Hi Bunuel

Could you please show how to go about working on the same problem if the ORDER of the dice mattered.
What would the approach be?

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

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28 Apr 2014, 21:22
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Hi Bunuel

Could you please show how to go about working on the same problem if the ORDER of the dice mattered.
What would the approach be?

Order of the dice matters means the dice are distinct. Think of having 3 dice of 3 different colors - Red, Blue and Yellow. So (5, 1, 1) is not 1 case now but 3 cases since Red die could show 5 or Yellow could show 5 or Blue could show 5.

Now its like 3 distinct places each of which can take 6 distinct values (1/2/3/4/5/6). So total number of outcomes = 6*6*6 = 216
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Re: How many different combinations of outcomes can you make by [#permalink]

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29 Apr 2014, 11:10
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If some of you are like me and find it difficult to think in combinations, I started by thinking about the possible outcomes, in order.

111, 112, 113, 114, 115, 116 = 6 possibilities

121 has already been counted with 112, so we notice that every digit goes in increasing order. Thus the next row has to start with 122 instead of 121.

122, 123, 124, 125, 126 = 5 possibilities.

Continuing on, we notice that our total number can be given by:

Starts with 1: 6+5+4+3+2+1 possibilities (111 - 116, 122-126, 133-136,....,166)
Starts with 2: 5+4+3+2+1 possibilities (222-226, 233-236, 244-246, 255-256, 266)
Starts with 3: 4+3+2+1 possibilities (333-336, 344-346,355-356,366)
Starts with 4: 3+2+1 possibilities(444-446, 455-456,466)
Starts with 5: 2+1 possibilities (555-556, 566)
Starts with 6: 1 possibility = 666

Thus, our total will just be the sum of 1+3+6+10+15+21 = 56.

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

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17 May 2014, 03:54
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.

Dear Bunnel

SO basically, when the qs stem says different combination outcomes when order doesnt matter it means that we have to take all possibilities in account:

XXX: i understood: 111, 222, 333, 444, 555, 666
XXY: what about XYX & XXY??? we have not counted this as qs says order doesnt matter?
XYZ: we have just counted 6*5*4/3! as order doesnt matter??

How do you get to know what the qs is asking?
when i read the qs I thought it was asking for all possible outcomes when three dice are thrown. so my answer was 6*6*6 = 216.
what is the difference between order doesnt matter and 216?
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Re: How many different combinations of outcomes can you make by [#permalink]

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17 May 2014, 05:09
NGGMAT wrote:
Bunuel 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 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.

Dear Bunnel

SO basically, when the qs stem says different combination outcomes when order doesnt matter it means that we have to take all possibilities in account:

XXX: i understood: 111, 222, 333, 444, 555, 666
XXY: what about XYX & XXY??? we have not counted this as qs says order doesnt matter?
XYZ: we have just counted 6*5*4/3! as order doesnt matter??

How do you get to know what the qs is asking?
when i read the qs I thought it was asking for all possible outcomes when three dice are thrown. so my answer was 6*6*6 = 216.
what is the difference between order doesnt matter and 216?

Yes, since the order of the dice does not matter then XXY, XYX and YXX are the same for this problem. Similarly, XYZ, XZY, YXZ, ... are also the same.
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Re: How many different combinations of outcomes can you make by [#permalink]

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17 May 2014, 06:14
Bunuel wrote:
NGGMAT wrote:
Bunuel 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 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.

Dear Bunnel

SO basically, when the qs stem says different combination outcomes when order doesnt matter it means that we have to take all possibilities in account:

XXX: i understood: 111, 222, 333, 444, 555, 666
XXY: what about XYX & XXY??? we have not counted this as qs says order doesnt matter?
XYZ: we have just counted 6*5*4/3! as order doesnt matter??

How do you get to know what the qs is asking?
when i read the qs I thought it was asking for all possible outcomes when three dice are thrown. so my answer was 6*6*6 = 216.
what is the difference between order doesnt matter and 216?

Yes, since the order of the dice does not matter then XXY, XYX and YXX are the same for this problem. Similarly, XYZ, XZY, YXZ, ... are also the same.

and when would it be 6*6*6??
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Re: How many different combinations of outcomes can you make by   [#permalink] 17 May 2014, 06:14

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