there are about 4 56s in the # 225
1
6! = 6*5*4*3*2*1 = 720
6! = 720
1-1 1-2 1-3 1-4 1-5 1-6 2-1 2-2 2-3 2-4 2-5 2-6 3-1 3-2 3-3 3-4 3-5 3-6 4-1 4-2 4-3 4-4 4-5 4-6 5-1 5-2 5-3 5-4 5-4 5-6 6-1 6-2 6-3 6-4 6-5 6-6 So there ARE 36 possible outcomes, you see. Answer BY: Magda Krysnki (grade sevener) :P
Between 5 and 6 hours
If you can repeat the numbers, it will be 6*6*6=216. If you can't, it will be 6*5*4=120.
One way to do this is to look at the probability of rolling a 7. Let's call that P(7) and the probability we seek is 1-P(7) Now we look at how many ways there are to get 7 with two dice 1+6 or 6+1 4+3 or 4+3 5+2 or 2+5 The sample space is 36 outcomes and there are 6 way to get 7 so 6/36=1/6 So the probability that the sum is 7 is 1/6 and 1-1/6=5/6 The answer is 5/6 ( my guess it that is higher than one might guess? You could also do the problem directly. Let's look at how to NOT get a sum of 7 1+2 or 2+1 1+3 or 3+1 1+4 or 4+1 1+5 or 5+1 2+3 or 3+2 2+4 or 4+2 2+6 or 6+2 3+5 or 5+3 3+6 or 6+3 4+5 or 5+4 4+6 or 6+4 5+6 or 6+5 6+6 5+5 4+4 3+3 2+2 1+1 There are 30 of these so 30/36 is 5/6
1000000 is 10^6, so it's 2^6*5^6, which has 7*7=49divisors
In a line in 6! = 6*5*4*3*2*1 = 720 ways.
2 technical fouls and 6 fouls
First, you add up 3, 6, 5, 4, and 2. Which is 20. Then, you had 5 numbers, so divide 20 by 5. Which equals 4. That is your mean, which is basically an average of many numbers.
2 and a half cups