Combinatoric discussion

The answer is 7!6! ways (about 3.63×10⁶)

Explain: Because of the same number of English and American people. So the English and American people had to sit 1-1 alternately.

I set the first Englishmen position. The remaining ones are then viewed as an arrangement on a straight line.

Permutate the remaining 6 Englishmen on their positions for 6! ways and 7 Americans for 7! ways. So, there is a permutation of all people 7!6! ways.

Problem 7:

In a solid box there is set of white and black chess pieces, each color contains 8 pawns, 2 knights, 2 Bishops, 2 Rooks, 1 queen and 1 king. Find the probability of these events.

(1) Randomly pick 1 piece and get a minor piece.

(2) Randomly pick 2 pieces at the same time and get 2 pawns of different colors.

(3) Randomly pick 2 pieces one by one by picking and putting them back and get both minor and major pieces are of the same color.

Note: bishop and knight are minor pieces, rook and queen are major pieces.

Notice that the first pick It is not determined whether to pick white or black, so the probability of the first pawn is 16/32.

The probability of picking both times according to problem 2 is 8/31 × 16/32 = 4/31.

GREAT!

Problem 8:

6 dices are rolled at the same time, find the probability that all dices will have different points.

Would it be like going 6/6*5/6*4/6*3/6*2/6*1/6?

It seems to me to be so because the probability for each one not being a previously checked number decreases by 1/6.

If so, it’s 1.543%

Yes!

number of event = 6!

number of sample space = 6⁶

So pobability is 6!/6⁶ = 5/324 ~ 0.0154

Problem 9:

There are 9 different books, how many ways can be divided into 3 piles, 3 books each?

It seems to me that a factorial would suffice. Although, if you want to consider each pile as it’s own entity and shuffle those around idk how to do that. Maybe you would just multiply it by 3 factorial.

Regardless, if my way is correct, the answer would be 362,880. If you want to include the 3 factorial part, it would be 2,177,280.

If this problem is to find the number of permutations of all books, the answer is 9! you are correct

However, this problem is only focused on dividing the book into sections, 3 sections, ignoring internal permutations.

What’s the new problem?

Almost right, since each pile had 3 equal books, there was no difference. So permutation of piles A,B,C does not occur. That makes the way to divide the group into 1680/3! = 280

Choose a domain from set A by n(A) ways.

Choose a range from set B by n(B) ways.

So, the total number of relations is n(A)×n(B).

Problem 11:

Given the relative universe 𝕌 = {1,2,3,4} and A, B are subsets of 𝕌

If S = { (A,B) | n(A∩B)=2 } then find the number of elements of S.

Thank you. I've always misunderstood the definition of "relation".

Relation from A to B is a set of ordered pairs (a, b) such that a ∈ A and b ∈ B.

The answer is 14, see the solution here.

https://www.chess.com/clubs/forum/view/combinatoric-discussion?page=2#comment-83592549

Problem 12:

Round table with 5 seats requires one type of food and drink to be served at each seat.

There are 5 different types of food and the drinks have the same 3 glasses of water and the same 2 glasses of tea, how many ways are there to serve the food and drink?

Hint : Consider the placement of the two tea cups. Then notice the rotation of the food position while maintaining the drink position. It will make the sorting format all different.

Problem 13:

There are 3 pairs of socks in a box. Randomly pick 2 socks. Find the probability that both socks are the same pair.

Problem 14:

How many ways can 4 couples sit around a round table? When each couple must sit next to each other.