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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 28 Jul 2018, 21:47
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.


I understand why you divide 6x5x4 by 3!. But why wouldn't for the same reason you divide 6x5 by 2! for the same reason?

Bunuel could you please explain this


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P and C  [#permalink]

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New post 24 Oct 2018, 06:33
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
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Re: P and C  [#permalink]

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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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P and C  [#permalink]

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New post 24 Oct 2018, 06:53
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



I think the correct answer is 56.

Here is the approach.

Case 1: All three dices have different numbers = 20C3 = 20( Its 20 C3 since order of dice is not important)
Case 2: Two dice has same number while third has different = 6*1*5 =30
Case 3: All have same number = 6*1*1 (We have six options for dice A and if that comes second dice and third dice should have same number(one option) so 1*1)


So total = 20+30+6 =56


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Re: P and C  [#permalink]

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New post 24 Oct 2018, 07:22
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


OA: C

Case 1 : When 3 dice show distinct numbers : \(C(6,3) = \frac{6!}{3!3!}=20\)
Case 2 : When 1 dice show distinct number and other 2 show same number : \(6*1*5 =30\)
Case 3 : When all 3 dices show same number : \(6*1*1 =6\)

Total number of case : \(20+30+6 =56\)
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Re: How many different combinations of outcomes can you make by  [#permalink]

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New post 25 Oct 2018, 16:54
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 &nbs [#permalink] 25 Oct 2018, 16:54

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