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

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01 Jul 2014, 05:31

Dear Bunuel, I could not understand one thing. For the case of XYZ----> the number of possible outcomes could be 6*5*4 . Why it is C(6,3)? Whats the difference between these two? Kindly explain.

Dear Bunuel, I could not understand one thing. For the case of XYZ----> the number of possible outcomes could be 6*5*4 . Why it is C(6,3)? Whats the difference between these two? Kindly explain.

Notice that C(6,3) = 4*5*6, so there is no difference between these two. C(6,3) gives the number of 3 different numbers out of 6, which is exactly what we want for XYZ.
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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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16 Dec 2014, 05:36

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.

Can somebody please explain ..why we didnt divide Case (2) by 2!..because it will also show repititions like Case (3). I cant understand the explanation given above

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13 Oct 2016, 20:03

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.

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

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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

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

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

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01 Nov 2017, 14:37

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,

Shouldn't the question stem be edited to "no. of 3-digit Combinations possible from three dice rolls" instead of just stating "different combinations possible from three dice rolls"?

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 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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13 Nov 2017, 17:58

Bunuel, do you think it is a good idea to struggle through some of these types of problems in our heads (without paper and pencil) in order to develop better math intuition? Maybe when we are falling asleep or something? I think some of these problems require number sense awareness of patterns that non-math experts like me haven't noticed. If we can get better at intuitively understanding some of the patterns, then maybe we won't make silly mistakes? This is especially true if we have to solve these problems in under 2 minutes. Then we can get even faster at solving problems on paper?