4 Dice

It would be easier to figure out the total number of outcomes and subtract that by the number of outcome where there are NO 6's.
Use the fundamental counting theory

total= (6)(6)(6)(6)=1296
outcomes with no 6's= (5)(5)(5)(5)=625

1296-625=671

Answer is D

Guys I've doubt. Although it is not explicitly mentioned over here that all dices are not identical, but aren't we suppose to take all the dices as identical???

(D) 671

6^4-5^4

= (6^2-5^2)*(6^2+5^2)

= (6+5)(36+25)

= (11)(61)

= 671

IMO D , 6^4-5^4=671

The total number of outcomes will be 6^4 here.

so we have 4 places (4 rolls of the dice) and numbers 1,2,3,4,5 can be used to fill these 4 places

if I take that as 5C4 = 5 ways. Then these 4 selections will move around among themselves so it will be 5* 4! ways = 120 ways. How do I get ahead from here ?

asciijai wrote:The total number of outcomes will be 6^4 here.

so we have 4 places (4 rolls of the dice) and numbers 1,2,3,4,5 can be used to fill these 4 places

if I take that as 5C4 = 5 ways. Then these 4 selections will move around among themselves so it will be 5* 4! ways = 120 ways. How do I get ahead from here ?

You have two options.

1:: Find the number of INVALID possibilities, then subtract that from the total.

You saw that the total is 6�. You will get NO 6s in 5� possibilities. So you're left with 6� - 5� possibilities in which there is at least one six. This gives 6� - 5�, or (6² + 5²)(6² - 5²), or 61 * 11, or 671.

2:: Find the number of VALID possibilities, case by case.

Exactly one 6 = 1 * 5 * 5 * 5 * 4 (since any of the four dice could be the 6)
Exactly two 6s = 1 * 1 * 5 * 5 * 6 (since any two of the four dice could be 6, and (4 choose 2) = 6)
Exactly three 6s = 1 * 1 * 1 * 5 * 4 (since any three of the four dice could be 6, and (4 choose 3) = 4)
All 6s = 1 * 1 * 1 * 1

This sum is 4*125 + 6*25 + 20 + 1, or 671.