Problem Solving

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

Pls explain.

Yikes...it took me almost 20+ minutes to solve this. There is no way I was gonna solve this on actual test....

Anyways, here is my chain of thought..

We basically have to consider 3 situations

A) When a number is repeated 3 times. (6,6,6) (5,5,5) etc

= 6 * 1 * 1

= 6

This makes sense as there are 6 numbers and if all the dice show the same number, we are choosing 1 number that will be repeated out of 6 choices. You could also use the formula for 6choose1 and you would arrive at the same thing.

B) When a number is repeated 2 times (6,6,5) (5,5,6) etc
= 6 * 1 * 5

= 30

This makes sense as we can have 6 choices for the 1st dice, 2nd dice has to be the same as the first one, and that leaves us with 5 numbers to choose from for the third choice since the number on the 3rd dice has to be different from the other 2.
This can also be considered as 6Choose1 (for the number that is to be repeated) * 5Choose1 (for the number that is not repeated). Which will give you 6 * 5.

C) When a number is not repeated (1,2,3) (6,5,4) etc

= 6!/(3!*3!)
= 5 * 4
= 20

Now this would be 6Choose3. As you are choosing 3 different numbers from choice of 6 possible numbers.


Now adding A + B + C = 6 + 30 + 20 = 56

Hence the answer is C

Hope this helps. Let me know if you still have questions/concerns.

crackgmat007 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

Pls explain.

Nice explanation. Tx.

I agree that this is hell of a problem to solve in 2 mins on the G day.

hard to believe that u got just 42 in quants on the gmat with that explanation!

What happened?

Thanks. The only possible explanation I can give is that when they say the penalty of not completing is huge...they really mean it is HUGE!! I did not finish the last question...

I have consistently scored over 90 percentile on all the practice tests. I have taken at least 10 practice tests and the lowest I have scored is 85 percentile i think....but the test day was a different story all together.

I am retaking it a few days...hope it is a little different this time.

Thanks.

hedonist123 wrote:hard to believe that u got just 42 in quants on the gmat with that explanation!

What happened?

Ankit, I have a question on the third calculation.

C) When a number is not repeated (1,2,3) (6,5,4) etc

= 6!/(3!*3!)
= 5 * 4
= 20

here since the order does not matter.
The number i am getting is 10.

if you consider (1,2,x) x can have 4 values.
similarly, (1,3,x) x can have 3 values
(1,4,x) x has 2
(1,5,x) x has 1

Everything else is a repeat since the order does not matter.
So this should give us 20.
Correct me if I am missing something here.

Thats a fair concern.

Here is what you have listed

ben wade wrote: The number i am getting is 10.

if you consider (1,2,x) x can have 4 values.
similarly, (1,3,x) x can have 3 values
(1,4,x) x has 2
(1,5,x) x has 1

This gives you 10 combinations...

But if you notice, All your examples have 1 as the first choice...

consider your choices without 1 being included...

(2,3,x) -> 3 combinations
(2,4,x) -> 2 combinations
(2,5,x) -> 1 combination

This gives you 6 more choices..

Similary if you start with 3 it will give you 3 more combinations..
and if you start with 4 it will give you 1 more combination...


Summing these will give you 20...



Cheers.

ankitns wrote:
C) When a number is not repeated (1,2,3) (6,5,4) etc

= 6!/(3!*3!)
= 5 * 4
= 20

Now this would be 6Choose3. As you are choosing 3 different numbers from choice of 6 possible numbers.

[/quote]

Shouldn't the above be simply--- 6x5x4= 120? Why 6C3? Please explain.

Thanks.

nipun_jindal wrote:

ankitns wrote:
C) When a number is not repeated (1,2,3) (6,5,4) etc

= 6!/(3!*3!)
= 5 * 4
= 20

Now this would be 6Choose3. As you are choosing 3 different numbers from choice of 6 possible numbers.

Shouldn't the above be simply--- 6x5x4= 120? Why 6C3? Please explain.

Thanks.

Hi Nipun!

The problem with treating this like a permutations problem - which is when you'd simply do 6*5*4 - is that we don't care about order, so you're counting certain combinations multiple times.

For example, the 3 dice could come up as a 5, a 4 and a 3. However, they could also come up as a 4, a 3 and a 5. You've counted these as two separate outcomes, even though the question explicitly says that order doesn't matter.

So, we need to get rid of duplications. Since there are 3! ways to arrange 3 objects, we divide by the extra 3! to eliminate the multiples. Accordingly, instead of :

6!/3! = 6*5*4

we get:

6!/3!3! = 6*5*4/3*2 = 5*4 = 20

Here's the takeaway: when order does NOT matter, you have to use the combinations formula; when order DOES matter, use the permutations formula instead.

Combinations: nCk = n!/k!(n-k)!
Permutations: nPk = n!/(n-k)!

I hope that clears up the confusion,

Stuart