probability

Question

6 dice are thrown simultaneously ,find the probability that exactly three of them shows the same face and remaining three shows different faces.

Answer 1

Total number of possibilities :  Total_Outcome=6^6

Choose 3 dice among the six for showing the same face

selections= 6C3

Common face can take 1,2,3,4,5,6

selections= 6C3 * 6 

Remaining 3 dice can take 5*4*3 

selections= 6C3 * 6*5*4*3

Probability = Selections/ Total_Outcome

Answer 2

6 dices thrown

3 of them showing 1 face out of all 6 faces of dice

other 3 can show any faces of dice among rest 5 faces of dice

So, probability will be

=$\frac{6C3*6C1+3C3*5C3}{6^6}$

Comments

Here it is written that exactly 3 of them shows same face,

now as per your conclusion, say 3 dices are showing face 6.

remaining 3 can show any of remaining 5 faces.

now lets say remaining 3 shows 5 all. then ...the statement exactly 3 of them shows same face is violated.

so i guess for remaining three dices selection should be like

5c1,4c1,3c1     .

.

.

correct me if i am wrong

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Your point is correct. But remaining 3 can show two faces same.

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Yes but in the worst case 3 faces can be same in the point discussed in the answer.. so i am getting doubt .. in mine no faces can be same... but point out the question remaining faces shows differen faces... so it may be the case that all are different

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yes, agree. Question seems like two faces same is not allowed.

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exactly three of them shows the same face and remaining three shows different faces

I think it means no 4th value will show same face

means (1,1,1),(3,3,3) allowed here

but (1,1,1,1) not allowed here

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P= $\frac{6* 5 * 4 * 3}{6^{6}}$