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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.
_________________

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;

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

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08 Sep 2013, 22:36

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.

Answer: C.

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;

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.

Answer: C.

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,

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

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17 Apr 2014, 02:54

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.

Answer: C.

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) \(C^2_6\)?

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 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.

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

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18 Apr 2014, 02:23

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.

Answer: C.

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.

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?

Thanks in advance.

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
_________________

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

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17 May 2014, 04:54

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.

Answer: C.

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?
_________________

Hope to clear it this time!! GMAT 1: 540 Preparing again

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.

Answer: C.

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.
_________________

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

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17 May 2014, 07:14

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.

Answer: C.

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??
_________________

Hope to clear it this time!! GMAT 1: 540 Preparing again