# Thread: Probability and counting problems

1. ## Probability and counting problems

I need help to solve these problems :

1. A dartboard is divided into 6 sections. These 6 sections is given numbers : 49, 5, 54, 28, 37, 23. What is the least number of throws needed in order to get a score of exactly 100?
I have noticed that a sum of 100 can be from the sum of 3 numbers : 23+23+54, 49+28+23. But I have no idea how to work it out.

2. In a party, there are 2006 cakes. The first group eats 1/2 of them. The second group eats 1/3 of the rest. The third group eats 1/4 of the rest. An so on .... The 2005th group eats 1/2006 of the rest. How many cakes have been eaten ?

2. Hello, cdesiani!

You've solved the first one . . .

1. A dartboard is divided into 6 sections.
These 6 sections is given numbers: 49, 5, 54, 28, 37, 23.
What is the least number of throws needed in order to get a score of exactly 100?

I have noticed that a sum of 100 can be the sum of 3 numbers: 23 + 23 + 54.
. .
So you found a three-number sum . . . Good!
A sum of 100 cannot be produced by one throw.
And it can be shown that a sum of 100 cannot be produced by two throws.
. .
[Just check out the 15 two-number sums.]

Therefore, at least three throws are needed . . . That's it!

3. 2) None, we have;

$r~=~\left(\frac{1}{2!}\right)+\left(\frac{1}{3!}\r ight)+\left(\frac{1}{4!}\right)+$ ...... $+\left(\frac{1}{n!}\right)$

As $n\rightarrow\infty~~r\rightarrow0.7182....$

I think

4. Hello, Sean!

You are on the right track . . .

. . $r \:= \: \frac{1}{2!}+ \frac{1}{3!}+\frac{1}{4!}+ \hdots + \frac{1}{2006!}$

We have the infinite series: . $e^x\;=\;1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \hdots$

. . If $x = 1$, we have: . $e \;=\;1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \hdots$

. . Then:. . $e - 2 \;=\;\frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \hdots$

Hence: . $r \;\approx\;e-2 \;=\;0.718281828459045\hdots$

Therefore: . $2006(e-2) \;\approx\;1440\frac{7}{8}$ cakes were eaten.

5. ## confused ....

Sorry, but what is e that is equal to 0.7182818284...?

Still