Melody

If the letters are different then I think it should be the same.

To work it out is much more difficult though.

There are 4 mailboxes and 4 letters, they are randomly placed in mailboxes, what is the probability that one of the mailboxes has exactly 3 letters.

It sounds to me like the letter boxes are different but the letters are effectively the same 

First I will look for the total number of ways

Here are the 4 letters     * * * *

We have to split those into 4 piles where some piles can have no letters.  

We need to seperate them using 3 bars

eg     ||*|***

that would be  non in the first box, none in the second, 1 in the third and 3 in the 4th.

So that would be  7C3  or 7C4 ways   =  35

NOW how many of those have 3  in one box   If there are 3 in the first box then there can be  1 in any of the othere 3 that is 3 ways

There are 4 letter boxes so that is  3*4 =12 ways

So I get    12/35

NOTE: This is called stars and bars method.

Think!

This is not a hard one.

Assumig peope are the same an rotations are the same

then

tie each person in a chair  to a chair  on the right,  

so now there are 5 double people and 8 other seats that is 13 altogether

Pur one double in place, down't matter where, that takes care of rotations

now there are    12C4 = 495 ways

are the people classed as identical?

are rotations the same?

Hi DaKaden

\(acute\;angle\;\beta=asin(\frac{b}{c})\\ so\\ obtuse\;angle\; B =180-asin(\frac{b}{c})\\~\\ acute\;angle\;\alpha=asin(\frac{a}{c})\\ so\\ obtuse \;angle\;A=180-asin(\frac{a}{c})\\\)

So area of obtuse triangle off B 

   \(=0.5*ac*sin(180-asin(\frac{b}{c}))\\ =0.5*ac*sin(asin(\frac{b}{c}))\\ =0.5*ac*\frac{b}{c}\\ =\frac{ab}{2} \)

Check what i have done very carefully, there is plenty of room for careless error.

You can find the area of  obtuse triangle off A.

find c in terms of a and b using Pythagoras's theorum.

You should be able to finish from there.

At a meeting, five scientists, two mathematicians, five journalists, two biologists, and three politicians are to be seated around a circular table.  How many different arrangements are possible if the scientists must all sit together (in five consecutive seats).  (Two seatings are considered equivalent if one seating can be obtained from rotating the other.)

5 scientist                (all together)

2 mathematicians

5 Journalist

2 biologists

3 politicians

17 people 

There are 5! ways to seat the scientist then  12! ways to seat everyone else

5! * 12! =  a LOT of ways.

\( (ab + ac + bc)^3 - a^3*b^3.\\ = (ab + ac + bc)^3 - (ab)^3\)

difference of 2 cubes

\(\boxed{a^3-b^3=(a-b)(a^2+ab+b^2)}\)

You can take if from there.

I'd like to see this one solved too (with plenty of explanation).   

Anyone interested?   

Right triangle XYZ has legs of length XY = 12 and YZ = 6.  Point D is chosen at random within the triangle XYZ.  What is the probability that the area of triangle XYD is at most 20?

If point D is anywhere in the region    XYAC  then the area of  XYD will be at most 20.   Can you work out why?

So the prob that XYD<=20    is          Area of XYAD divided by  Area of  XYZ