Combinatoric discussion

Your solution is correct!

However, in this problem, 4 couples means 4 male and 4 female, for a total of 8 people.

When there are 8 people, the number of ways to sit increases until it is difficult to count.

HINT: Find the number of ways to arrange the positions of the 4 couples first. Then each couple can switch places with each other in 2 ways.

Solution:

Select 2 elements from 𝕌 that are in A∩B can be done in ⁴C₂ = 6 ways.

Arrange the remaining two elements, each of which can be in set A only, set B only, or not in both sets. There are 3² = 9 ways.

So, There are (A,B) corresponding 6×9 = 54 ordered pairs.

Yes, The solution is easy to understand but taking the initiative is not easy.

Solution:

Considering only the arrangement of drinks on the circle, We can do this in only 2 ways. (look at the distance between the two cups of tea)

Then 5 foods are then viewed as arranged on a straight line. (Because the rotation changes the position of the food relative to the drink), there are 5! = 120 ways to do it.

Therefore, there are a total of 2×120 = 240 ways to serve food and water.

Problem 15:

Find the number of rectangles within a 6×3 grid.

Is there a short way to do this, or do we have to go through each case of the rectangle? Please tell me there’s a short way, It’d be much more fun to find

Hint: Select two points on the wide side of the grid and two points on the long side of the grid. to create various rectangles.

Problem 15-1:

Find the number of rectangles within a 3×2 grid.

This is a study case of the latest problem. You can count directly.

Then I will present a faster method. which you can use to solve problem 15.

A square is a (special kind of) rectangle.

It looks like you didn't include the squares.

The rectangle refers to a quadrilateral with all angle is a right angle. So this will include the square.

That's right! Here's how to calculate it without counting.

The 3 unit lenght side of the grid has 4 vertices. Choose 2 vertices to form different horizontal interior patterns. can do in ⁴C₂ = 6 ways.

The 2 unit width side of the grid has 3 vertices. Choose 2 vertices to have different vertical interior patterns. can do in ³C₂ = 3 ways.

So, there are 6×3=18 different rectangles as this diagram.

Can you use the same method to find the number of rectangles in a 6×3 grid? try it.

Oops! It should be (7²-7)/2 and (4²-4)/2

Because ⁷C₂ = 7!/(5!2!) = 7×6/2 = (7²-7)/2

Mate!

Problem 16:

How many 3-digit numbers are divisible by 5 or 7?

There are 2 errors.

1. 3-digits numbers start from 100 and end at 999
2. Some numbers are divisible by both 5 and 7 (divisible by 35). When the quantity of numbers that are divisible by 5 is combined with quantity of numbers that are divisible by 7, these numbers will be counted twice.

Therefore, the solution is to add the answers obtained in both cases and subtract the quantity of numbers that are divisible by 35.

Solution:

Let aₙ represent the set of 3-digit numbers that are divisible by n.

a₅ = {100, 105, 110, ..., 995} ; n(a₅) = ((995-100)/5)+1 = 180

a₇ = {105, 112, 119, ..., 994} ; n(a₇) = ((994-105)/7)+1 = 128

a₃₅ = {105, 140, 175, ..., 980} ; n(a₃₅) = ((980-105)/35)+1 = 26

So n(a₅∪a₇) = n(a₅)+n(a₇)-n(a₃₅) = 180+128-26 = 282

Problem 17:

There are 4 different blue pens and 3 identical red pens. Find the number of ways to carry at least one pen.

Problem 18:

How many 4-digit integers are there that have only two different number in each digit?

ex. 1414, 6636