COUNTING

 

Fundamental Concepts and Techniques

Counting is a crucial part of math that arrangements with deciding the quantity of components in a set. It frames the reason for further developed ideas in likelihood, measurements, and combinatorics. This guide will present essential counting standards and procedures, including the central counting rule, changes, and mixes.The Fundamental Counting Principle

The Fundamental Counting Principle states that if there are $n$ ways to do one thing and $m$ ways to do another, then there are $n \times m$ ways to do both. This principle can be extended to more than two events.

For example:

• If you have 3 shirts and 4 pants, the total number of possible outfits is: $$3 \times 4 = 12$$

Permutations

Stages are utilized when we are counting the quantity of ways of orchestrating a bunch of things where the request matters. The quantity of changes of n particular things is given by 𝑛!

n! (n factorial), which is the result of all sure whole numbers up to $n$: $n! = n \times (n-1) \times (n-2) \times \ldots \times 1$

For example:

• The number of ways to arrange 3 books on a shelf is: $$3! = 3 \times 2 \times 1 = 6$$

When arranging a subset of $k$ items from $n$ distinct items, the number of permutations is given by: $P(n, k) = \frac{n!}{(n-k)!}$

For example:

• The number of ways to arrange 2 books out of 4 on a shelf is: $$P(4, 2) = \frac{4!}{(4-2)!} = \frac{4 \times 3 \times 2 \times 1}{2 \times 1} = 12$$

Combinations

Combinations are used when we are counting the number of ways to choose a subset of items where the order does not matter. The number of combinations of $n$ items taken $k$ at a time is given by: $C(n, k) = \binom{n}{k} = \frac{n!}{k!(n-k)!}$

For example:

• The number of ways to choose 2 books out of 4 is: $$C(4, 2) = \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3 \times 2 \times 1}{2 \times 1 \times 2 \times 1} = 6$$

Example

Question No.01:

In bridge, the 52 cards of a standard deck are dealt to four players. How many different ways are there to deal bridge hands to four players.?

Solution:

Combinations refer to the selection of a specific number of items from a larger group, where the order of selection doesn't matter. In this case, we need to select 13 cards (hand) from the 52 cards (deck) for each of the four players.

If 52 cards are dealt to four players, then each of the four player will receive 52/4​=13 cards

Using the formula for combinations, the number of ways to deal bridge hands to four players is:

C(52, 13) * C(39, 13) * C(26, 13) * C(13, 13)

where C(n, r) represents the number of combinations of r items chosen from n distinct items.

Evaluating this expression, we get:

52! / (13! * 39!) * 39! / (13! * 26!) * 26! / (13! * 13!) * 13! / (13! * 0!)

Simplifying further, we have:

C=52! / (13!)^4

This expression represents the total number of possible hand distributions in bridge, considering each player receives 13 cards from a standard deck of 52 cards.

 

 

Question No.02:

 A bowl contains 10 red balls and 10 blue balls. A woman removes balls at random without looking at them. How many balls must she remove to be sure of having at least three balls of the same color?

Solution:

In order to ensure that she has 3 balls of the same color after 5 pickups, we can consider the worst-case scenario where she alternately picks up balls of different colors. Let's assume there are two colors, say Red and Blue.

After the first pickup, she has 1 Red and 0 Blue. After the second pickup, she has 1 Red and 1 Blue. After the third pickup, she may have (2,1) or (1,2). After the fourth pickup, she may have (2,2). After the fifth pickup, she may have (5,0), (4,1), or (3,2).

Now, applying the pigeonhole principle: She has 2 pigeonholes (the colors Red and Blue) and 5 picks. By the pigeonhole principle, at least one pigeonhole must contain at least [5/2] = 3 balls. Therefore, in the worst case, she is guaranteed to have 3 balls of one color after 5 pickups.

Question No.03:

A bowl contains 10 red balls and 10 blue balls. A woman removes balls at random without looking at them. How many balls must she remove to be sure of having at least three blue balls?

Solution:

The worst-case scenario is that the first 10 balls are red. In order to be sure she has 3 blue balls, we need to consider the pigeonhole principle. Let's consider two pigeonholes: Red and Blue. The worst-case scenario is that the first 10 balls are red (all in the Red pigeonhole). Now, for the next 3 pickups, she will guarantee to pick at least 3 blue balls.

Therefore, in the worst case, using the pigeonhole principle, she needs to pick up 13 times to be sure she has 3 balls of the other color (Blue).ways

Question No.04:

 How many ways are there for a horse race with four horses to finish if ties are possible? (Note that since ties are allowed, any number of the four horses may tie.)

Solution:

If there are no ties the number of orders is 4! = 24

 

If 2 tie the number of orders is 4!/2! = 12

 

If 2 tie and the other 2 tie the number of orders is 4!/[2!*2!] = 6

 

If 3 tie the number of orders is 4!/3! = 4

 

If 4 tie the number of orders is 4!/4! = 1

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Total # of orders is 24+12+4+1 = 47