If 6 dice are rolled. How many outcomes are there
$(a)$ If all of the numbers are even.
$(b)$ if the numbers must appear in pairs.
What i try ::We have even numbers as $2,4,6.$
So number of ways in which even number appears in upper face of dice $=3\cdot 3\cdot 3\cdot 3\cdot 3 \cdot 3=3^{6}$
For second part
Means any numbers appera in pairs
We have $6$ numbers , So total ways
$=6\cdot 5\cdot 4=120$
Can anyone please explain me is i am right. If not Then How do i solve it, Help me , Thanks
As lulu pointed out, the problem doesn't clearly specify whether or not outcomes are distinct up to permutation. In this answer, I assume that different permutations are different outcomes.
$(a)$ Like you pointed out, there are exactly three choices for the even number; however, there are six total dice rolls, which means that the answer should be $3^{6} = 729$ rather than $3^{3}$ (there are $3$ choices for each of the $6$ rolls).
$(b)$ Let $x_1, x_2, \ldots, x_6$ denote the number of times that $1, 2, \ldots, 6$ appear in our dice rolls, respectively. By the problem constraints, we require $x_i = 0, 2, 4, \text{ or } 6$ and $\sum_{i} x_i = 6$. Note that we also need to take permutations into account (i.e. $(2, 2, 2, 4, 4, 4)$ is different from $(2, 4, 2, 4, 2, 4)$, but they both have $x_2 = 3, x_4 = 3$). We will perform casework.
First, consider the case in which some $x_i$ equals $6$. There are exactly ${6\choose 1} = 6$ positions for the $6$ to be placed in (no other values can be non-zero). This contributes $6$ to the answer.
Next, consider the case the case in which some $x_i$ equals $4$ and another equals $2$. There are exactly ${6\choose 2} = 15$ ways to pick the two numbers, and we can permute which $x_i$ equals $4$ and which equals $2$ in $2$ ways. Also, we can permute the order of the dice rolls in $6!/(4! \cdot 2!) = 15$ ways. This contributes $15 \times 2 \times 15 = 450$ to the answer.
Finally, consider the case in which three non-zero $x_i's$ equal $3$. There are exactly ${6\choose 3} = 20$ ways in which this can happen. Also, we can permute the sequence itself is $6!/2^3 = 90$ ways. This contributes $20 \times 90 = 1800$ to the answer.
Thus, the answer is $20 + 450 + 1800 = 2270$.