I don't have the exact answers for these questions, I only know whether the answers would be right or wrong, my answers below are wrong though
1.
solved
2.
[solved]
solved
5.
[solved]
6.[solved]!!
-------------------------
well I don't know which categorary Permuations and combinations fall under, sorry if I posted in the wrong forum. Any help would be greatly appreciated
Hello, kakab00!
There are: . ways.4. There are eight seats at a round table, equally spaced.
Find the number of ways in which two women, one married couple,
and one family of four people can be seated at the table
(a) if their are no restriction to which seats they must take.
Duct-tape the married couple together: .(b) if the married couple must be seated together
and the family of four must be seated together.
Duct-tape the family-of-four together: .
Then there are four "people" to arrange: .
. . and they can be seated in: . ways.
But for each of these seatings,
. . the married couple can be rearranged in ways
. . and the family-of-four can be rearranged in ways.
Therefore, there are: . possible seatings.
This is much trickier; I had to baby-talk my way through it.(c) if the married couple must sit together
and the two women must not be seated directly opposite each other.
Seat the married couple.. .Call their seats #1 and #2.
The diagram looks like this:Code:Husband [1] Wife [8] | [2] \ | / \ | / [7]- - - * - - - [3] / | \ / | \ [6] | [4] [5]
Now we will seat the two women:
Select one of the women; there are 2 choices.
Suppose she sits in chair #5 or #6; 2 choices.
Since the other woman can not be seated opposite.
. . then the remaining five people can be seated in 120 ways.
Hence, there are: . 480 seatings.
Select one of the women; there are 2 choices.
Suppose she sits in chair #3, 4, 7,or 8 . . . 4 choices.
. . Then the other woman must not sit in the opposite chair; she has 4 choices.
Then the family-of-four can be seated in 24 ways.
Hence, there are: . 768 seatings.
So there are: . 1248 seatings.
But originally, we placed Husband in chair #1 and Wife in chair #2.
. . Of course, they could be reversed.
This doubles the number of seatings.
Therefore, the number of seatings is: . 2496
Hello, kakab00!
By DeMorgan's Law: .The events A and B are independent.
If it given that: .
(a) What is the value of ?
So we have: .
. . Hence: .
Events A and B are independent.
.. Hence: .
Formula: .
So we have: . .
Then: .
Therefore: .
We know that: .(b) What is the value of ?
. . Hence: .
By DeMorgan's Law: .
Therefore: .
Since and , then: .(c) It is given that events B and C are mutually exclusive,
. . . and
What is the largest possible chance that event C will occur?
. . (Circle is completely contained in circle .)
Since and are mutually exclusive, their circles do not intersect.
The Venn diagram looks like this:Code:* - - - - - - - - - - - - - - - - - - - - - - - - - * | | | *---------* *---------* | | / A \ / B \ | | / × \ | | / 0.05 / \ 0.60 \ | | * * * * | | | | | | | | | *---* | | | | | | / \ | 0.20 | | | | | * * | | | | | | | C | | | | | | * * * * * * | | \ \ / \ / / | | \ *---* × / | | \ / \ / | | *---------* *---------* | | 0.15 | * - - - - - - - - - - - - - - - - - - - - - - - - - *
Since circle could be equal to circle
Hello, kakab00!
Don't delete your questions.
Others may benefit from the exchange we've had here.
I finally took the time to work out #3.
On side , there are 2 choices of points.(i) How many line segments are there joining any two points on different sides?Code:* * * o o C * o A * o o o * * * * * * o * o * o * * * * * * B
Each can be joined to any of the other 7 points.
. . There are: . line segments.
On side , there are 3 choices of points.
Each can be joined to any of the 4 points on side .
. . There are: . line segments.
Therefore, there are: . possible line segments.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Check
There are 9 points and there are: pairs of points.
But some of these join two points on the same side of the triangle.
There is 1 such pair on side .
There are such pairs on side .
There are such pairs on side .
. . Hence, there are: disallowed segments.
Therefore, there are: allowed line segments.
(ii) How many triangles can be formed from these points?
There are 9 points; we will select 3 of them.
. . There are: . sets of 3 points.
But some are sets of 3 collinear points, which do not form a triangle.
. . How many sets of three collinear points are there?
There is one set on side .
There are: sets on side .
. . Hence, there are: sets of collinear points.
Therefore, there are: . triangles.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Check
On side , there is 1 pair of points.
It can be joined with any of the other 7 points.
. . Hence, there are: triangles.
On side , there are: pairs of points.
They can be joined with any of the other 6 points.
. . Hence, there are: triangles.
On side , there are: pairs of points.
They can be joined with any of the other 5 points.
. . Hence, there are: triangles.
We can also pick one point from each side.
There are 2 choices for side , 3 for side , 4 for side .
. . Hence, there are: triangles.
Therefore, there are: triangles.