jfan17

Did you look at my steps? I got 10*9*2 because it matters in what order they select their dishes. You can switch around Yann and Camille; that was the rationale behind my multiplying by 2

Not too sure about this one, but I'll give it my best shot.

We know that There are 10 items on the menu, so say Yann goes first to order. She has 10 options to choose from, simple enough. Next, camille has 9 options to choose from, because they refuse to order the same dish, meaning she has 1 less option to order from. Then, we multiply this result by 2, because we can switch Yann and Camille around; if Camille goes first instead of Yann. We then get:

\(10*9*2 = 180\) ways of ordering dishes

I got an answer for you now that I've thought about it:

Realize that there are a total of 8 ways of winning that DON'T include the free space, and for each of these 8 ways, there's only one possible combination of these 5 numbers that let's corey fill it. There are 4 winning lines that DO include the free space, and each has 71 combinations of called numbers because it's equivalent to the four numbers in the non free spaces plus any of the leftover. The cumulative probability of corey's winning line also including the free space in the middle is equivalent to:

\(4*71/(8+4*71) = 71/73\)

First, the number of ways to choose 5 questions out of 15 is equivalent to 

\(15 \choose 5 \)= \(15!/(10!*5!) = 15*14*13*12*11/(5*4*3*2*1)\)

Canceling out top and bottom, we get:

\(3*7*13*11 = 3003\)

Next, we need the probability that he gets exactly five questions, which is:

\((1/2)^5\)

However, we haven't accounted for the next 10 questions which he needs to get wrong, which is a probability of 

\((1/2)^{10}\)

Our answer is then:

\(3003*{1/2}^5*{1/2}^{10} = 3003/(2)^{15}\approx 0.0916442871\)

Oh right, thanks for correcting me geno! I missed out on that small detail!

Hey guest! Glad to see you trying the problem. I skimmed it over though, and it appears it isn't as simple as finding how many ways there are to win. That's because for all we know, Corey could've won on her 11th or 12th turn(I don't know the exact number). The point is, we don't know on what turn she won on, meaning the cases are a lot more varied than you previously accounted for. 

Hey there guest! Glad to see you posting something about logarithms!(I need to brush up on them lol).

Let's get to the problem.

It's first given that f(n) is the base 10 logarithm of the sum of the elements in the nth row of pascals triangle. 

A key thing that you should realize is that the nth row of pascals triangle can be represented as: \({n \choose 0} + {n \choose 1} + {n \choose 2}.... {n \choose n}\)

Coincidentally, this also equals \(2^n\). You can try this out yourself with the first few rows of pascal's triangle!

The function f(n) then becomes:

\(f(n) = \log_{10}2^{n}\)

The expression we are asked to simplify then turns into:

\(\log_{10}(2^n)/ \log _{10} 2\)

Here, we have a neat log property we can use:

\(\log_{a} b / \log_{a} c = \log_{c} b\)

We then convert

\(\log_{10}(2^n)/ \log _{10} 2 = \log_{2}2^n\)

With the basic definition of a log, we then have

\(\log_{2}2^n = n\)

So the expression in terms of n is just n

Quick edit / note : Remember that the "first row" of pascal's triangle with just 1 term, is actually counted as the "0th" row.

First, we start with placing the 4x4 square so that one corner is (0,0), and the opposite corner is (4,4). We can define the x and y "coordinates" of Larry's square as x₁ and y₁. We define for laura's square x₂ and y₂ for the coordinates. That means that Larry's square and Laura's square will only ever intersect when the intervals of X1 and X2 intersect, and likewise for Y1 and Y2. 

Let X₁ equal to [a, a+1] and X₂ equal [b, b+1] with \(0 \leq a \leq3\) and \(0 \leq b \leq3\). This implies that the intervals given(x1 and x2) would only intersect in the case where |a - b| = 1(since that means the distance would be less than 1, which is the length of one of the unit squares. The region where the following holds true would be partitioned off into two 45-45 90 triangles, both with leg length of 2. The area of the desired region would be:

\(9-(2*2*1/2)*2 = 9-4= 5\)

Now, we have the probability that the intervals X₁ and X₂ intersect, which is \(5/9\). Likewise, the probability that Y₁ and Y₂ intersect is also \(5/9\), so our answer is just:

\((5/9)^2 = 25/81\)

I feel like I've done this problem somewhere before lol, can't seem to remember where

any time!