# Three fair six-sided dice are tossed and the numbers showing on top are recorded.

I don't know if they are correct, these are my attempts

Three fair six-sided dice are tossed and the numbers showing on top are recorded.

1. How many different record sequences are possible?
2. How many of the records contain exactly one six?
3. How many of the records contain exactly 2 four?

1) Since order does matter, Say for example , (1,2,1) is a different sequence then (2,1,1) Therefore this is P(6,1)*P(6,1)*P(6.1)

2) Now, order does not seem to matter. So, C(6,1)*C(6,1)*C(1,1) Something tells me this is terribly wrong

3) C(6,1)*(C,1,1)*C(1,1)

Edit:

2) C(3,1)*P(5,1)*P(5,1)

3) C(3,2)*P(5,1)

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(2) and (3) seem to be continuation of (1) so check that out with your source what exactly is being asked. –  ashley Dec 14 '12 at 23:41
I am somewhat concerned that you may be thinking too much in terms of symbols and formulas. Observe how concretely the people who answered approached the problem. –  André Nicolas Dec 15 '12 at 2:19

Hint: You are right for 1. For 2, you have three ways to pick which slot gets the six and five ways to pick the number in each of the other two slots. For 3, you have (how many?) ways to pick the slots for the two fours and (how many?) ways to pick the value of the other die.

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2) C(3,1)*P(5,1)*P(5,1) 3) C(3,2)*P(5,1) this correct? –  Aaron Dec 14 '12 at 23:43
@Aaron: that is correct. –  Ross Millikan Dec 14 '12 at 23:50

Your answer to (1) is correct, though it can be more simply expressed as $6^3$. You’re also right in thinking that you’ve gone astray in (2). Order does still matter: the $6$ can be in any of $3$ positions in the sequence, so there are $3$ ways to choose where it appears. Then each of the other two positions can be filled in $5$ ways (i.e., with anything but a $6$), so the total number of possibilities is ... ?

The reasoning needed for (3) is very similar to that needed for (2). How many choices are there for the position of the non-$4$ in the sequence? How many possible values are there for the non-$4$?

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(1) You have six choices for the first entry, six for the second, and six for the third. So $6^3$.

(2) Suppose a 6 is in the first slot. Then there's $5^2$ possibilities for the last two entries. On the other hand, 6 could have been the second entry, so there are $5^2$ possibilities for the first and third entries. Finally, there are $5^2$ possibilities for the first two entries if a 6 occurs in the last slot. So $3*5^2$.

(3) Reason as in (2), just think the two fours could occur in entries 1-2, 1-3, or 2-3.

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So the answer to 2) is just 3 * 6^2? –  Aaron Dec 14 '12 at 23:49
For 2: If the 6 is in the first slot, you are not allowed a 6 in either of the others. You have counted $661$ as acceptable, and in fact counted it twice. –  Ross Millikan Dec 14 '12 at 23:52
So i was right?!@#! –  Aaron Dec 14 '12 at 23:53
@RossMillikan: Yes, I mistyped meant a 5 there since we cannot repeat the 6 in the other slots. (Too many 6's floating around!) –  JohnD Dec 14 '12 at 23:56
@JohnD, is your answer identical to mine? –  Aaron Dec 14 '12 at 23:58