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.