Ps Rolled dices

Question

hello

from 4gmat software :

3 dices are rolled simultaneously in how many outcomes will the facing side of the 2 dices shows the same number 

plz explains your works and your reasoning ...


thanks

qpoo · Answer

I am ready to be the first one to get it wrong again, mandy ...

Is that 30 ?

for the first dice, you can choose any of the 6 numbers,
for the second one, you can only one of 5 remaining numbers,
the the third one, you can only choose one - the same as the second dice.

That makes it: 6x5x1 = 30

coffeeloverfreak · Answer

I know this is the long and not-too-elegant solution, but I fudged this out longhand and got 90 as the answer:

Ways in which 2 of the dice will show 1s (but not all 3):
112,113,114,115,116,121,131,141,151,161,211,311,411,511,611 = 15 ways.

Multiplied 15x6 because there are 6 numbers and got 90.

I know there's a way to do this with combination and permutations, but I'm too tired to work it out.

mandy · Answer

OA is 90 I will post the OA in a few days if needed

old_dream_1976 · Answer

Now that you gave us the answer I know the correct approach 

total outcomes = 6x6x6 = 216
outcomes with all distinct outcomes = 6x5x4 = 120
outcomes with all same numbers on the 3 dices = 6 

216 - 120 - 6 = 90

good one thank you

ps_dahiya · Answer

This is a nice question.

Total ways = 6* 6* 6 = 216

Ways when all three have same numbers = 6
Ways when none of them has same numbers = 6* 5 * 4 = 120

ANS = 216-6-120 = 90

Yurik79 · Answer

Good that we have you here)))Gr8 explanation

lhotseface · Answer

First, calculate the probability that a given number ( say 1 ) comes up on two of the dices 

1/6*1/6*5/6 ( 5/6 since the third number can be anything but 1 ).

Since there are 6 numbers we multiply it by 6. 

The result is (1/6*1/6*5/6)*6. Since for any given outcome with 2 dice having the same number, the outcome can be arranged in 3!/2! = 3 ways, We need to multiply that as well, 
(1/6*1/6*5/6)*6*3 = 90/216 is the probability. Thus, the total favourable outcomes are 90.

BTW, I like p_s_dahiya's method better. No need to involve prob...just count.

Ps Rolled dices

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