BuilderBoi

Marvin has to take 4 + 6 = 10 steps to get to his home, and there are \({10 \choose 4} = 210\) ways to do this. 

Now, we have to subtract the number of ways to get to (2, 4). There are \({6 \choose 2} = 15\) ways to do this. 

So, there are \(210 - 15 = \color{brown}\boxed{195}\) ways to do this.

\({1 \over a} + {1 \over b} = {b \over ab} + {a \over ab} = {a + b \over ab} = {7 \over ab} \)

\((a+b)^3 = a^3 + b^3 + 3a^2b+3ab^2 = 343\)

\(44 + 3a^2b+3ab^2 = 343\)

\(3a^2b+3ab^2 = 299\)

\(3ab(a+b) = 299\)

\(3ab \times 7 = 299\)

\(21ab = 299\)

\(ab = {299 \over 21}\)

\({7 \over {299 \over 21}} = {{147 \over 7} \over {299 \over 21}} = \color{brown}\boxed{147 \over 299}\)

What do you think? 

Is there a huge gap among the data points? 

IF you were given the first 3 points, do you think you could predict the 4th one? What about the 5th one?

There are \(6^4 = 1296\) ways to roll the dice. 

Use stars and bars to count the number of ways to roll a 12. Each roll must be at least 1, so we subtract \(4 \times 1 = 4\) to get 8 stars and \(4 - 1 = 3\) bars.

This makes for \({8 + 3 \choose 3} = {11 \choose 3} = 165\) ways to roll a 12. But, we can't have cases like 6, 1, 1, 0; 6, 2, 0, 0; 7, 1, 0, 0; and 8, 0, 0, 0. 

There are \({4! \over 2!} = 12 \) ways for each of the first 3, and there are \({4! \over 3!} = 4 \) ways for the last one, which sums to 40. 

So, there are 165 - 40 = 125 ways to roll a 12, which makes for a probability of  \(\color{brown}\boxed{125 \over 1296}\) 

There are \({10 \choose 5} = 252\) ways to choose the groups. 

If Jenny and Kenny are in the same group, there are 3 spots left, which makes for \({8 \choose 3} = 56\) ways to be in the same group. 

So, there are 252 - 56 = 196 ways for Jenny and Kenny to be in different groups. 

The probability is then \({196 \over 252} = \color{brown}\boxed{7 \over 9}\)

\({1 \over (a-1)} + {1 \over (b-1)} = {(b-1) \over (a-1)(b-1)} + {(a-1) \over (a-1)(b-1)} \)

\({(b-1) \over (a-1)(b-1)} + {(a-1) \over (a-1)(b-1)} = {a + b -2 \over {ab−a−b+1}}\)

\({a + b -2 \over {ab−a−b+1}} = {a + b -2 \over {ab−(a+b)+1}}\)

\(2x^2-7x+2 = x^2-11x+1\)

\(x^2 + 4x + 1 = 0\)

\(a + b = -{b \over a} = -4\)

\(ab = {c \over a} = 1\)

\({(-4) - 2 \over (1) - (-4) + 1} = {-6 \over 6} = \color{brown}\boxed{-1}\)

The product is different (i think) because to get an even product, you need even x even or even x odd; odd only happens with odd x odd. 

There are just as many ways to roll even as odd, so the answer is 1/2

At her average speed, she gets to work in \({30 \over 45} = {2 \over 3}\) of an hour or 40 minutes. 

To get to work 10 minutes earlier, she needs to go 30 miles in 30 minutes, or a speed of \({30 \text{ miles} \over 0.5 \text{ hours}} = \color{brown}\boxed{60 \text{ mph}}\)

Who's Mr.BB? Is it me?