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ESO Y4B · Mathematics
Combinatorics · Combinatorics

Introduction to combinatorics

How many different menus can you order by combining a starter, a main course and a dessert? How many passwords are there with 7 characters? Writing out the complete list by hand would be absurd. Combinatorics is the branch of mathematics that counts without having to list everything: in Bachillerato you will need it constantly, especially in probability. This block gives you the basic tools.

Before you start, make sure you know: Random experiments and sample space (2nd ESO)
Before you start, make sure you know: Concepts of probability (3rd ESO)

The multiplication principle

A restaurant offers 4 starters, 3 main courses and 5 desserts. To fix a complete menu you make one decision at each stage, and the decisions are independent of one another.

Block diagram representing the three independent stages of a menu (4 starters, 3 main courses and 5 desserts) giving rise to 60 combinations. 4 starters 3 main courses 5 desserts = 60 one decision per stage, independent of each other 4 · 3 · 5 = 60 different menus
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A restaurant offers a set menu with $4$ starters, $3$ main courses and $5$ desserts. How many different ways can a full three-course meal consisting of a starter, a main…

A restaurant offers a set menu with $4$ starters, $3$ main courses and $5$ desserts. How many different ways can a full three-course meal consisting of a starter, a main course and a dessert be ordered?

Solve
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A bicycle brand manufactures a model that is available in $3$ different frame sizes, $5$ colours and with $2$ different types of brakes. How many different versions of…

A bicycle brand manufactures a model that is available in $3$ different frame sizes, $5$ colours and with $2$ different types of brakes. How many different versions of this bicycle model are on the market?

Solve

Tree diagrams with dependent events

The multiplication principle also works when the number of options changes from one stage to the next, as long as you know how many remain on each branch. A tree diagram is the clearest way to see this.

Tree diagram representing two consecutive draws of balls without replacement, showing the reduction in options on the branches of the second stage. urn: 3 red, 2 blue — 2 are drawn without replacement red (3/5) blue (2/5) red: 1 r, 2 b left blue: 2 r, 1 b left red: 2 r, 1 b left blue: 3 r, 0 b left

Notice: after drawing a red ball, only 4 balls remain (2 red and 2 blue) for the second draw, because it is not replaced. The number of options in the second stage depends on what happened in the first, but the tree still has as many final branches as the product of the options at each level.

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Draw the tree diagram for the experiment consisting of tossing a fair coin and then rolling a tetrahedral die (with its faces numbered from $1$ to $4$). How many…

Draw the tree diagram for the experiment consisting of tossing a fair coin and then rolling a tetrahedral die (with its faces numbered from $1$ to $4$). How many possible outcomes does this compound experiment have?

You can do it on paper and send a photo with your answer.

Solve
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Two basketball teams, Alfa and Beta, play a best-of-$3$ playoff (the first team to win $2$ games qualifies). Draws are not possible. Draw the tree diagram with all…

Two basketball teams, Alfa and Beta, play a best-of-$3$ playoff (the first team to win $2$ games qualifies). Draws are not possible. Draw the tree diagram with all possible outcomes of the playoff. In how many different ways can the series of games unfold?

You can do it on paper and send a photo with your answer.

Solve
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An urn contains $3$ red balls and $2$ blue balls. Two balls are drawn sequentially without replacement. Draw the tree diagram for the experiment and calculate how many…

An urn contains $3$ red balls and $2$ blue balls. Two balls are drawn sequentially without replacement. Draw the tree diagram for the experiment and calculate how many distinct outcomes can be obtained if the order of extraction and the colour of the balls are taken into account.

You can do it on paper and send a photo with your answer.

Solve

The factorial

As soon as a task consists of placing elements one by one and there is one fewer available at each stage, the same product always appears: $n \cdot (n-1) \cdot (n-2) \cdots 1$. Because of its importance it has its own name and symbol.

Comparison table with the values of the factorials from 3 to 6, showing how their value grows rapidly from 6 to 720. 3! = 6 4! = 24 5! = 120 6! = 720 grows much faster than n: from 6! to 7! = 5 040 you just multiply by 7

When a large factorial is divided by a smaller one, you do not need to expand either of them completely: almost everything cancels.

The same idea works with letters: $\dfrac{(x+2)!}{x!} = (x+2)(x+1)$, because $(x+2)! = (x+2)(x+1) \cdot x!$.

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Calculate the value of the following numerical expressions using the properties of factorials: $\frac{8!}{6!}$ $4! + 3!$

Calculate the value of the following numerical expressions using the properties of factorials:

  • $\frac{8!}{6!}$

  • $4! + 3!$

Solve
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Express the following products using factorial notation: $15 \cdot 14 \cdot 13$ $7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$

Express the following products using factorial notation:

  • $15 \cdot 14 \cdot 13$

  • $7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$

Solve
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Simplify the algebraic expression $\frac{(x+2)!}{x!}$ and then solve the equation $\frac{(x+2)!}{x!} = 72$.

Simplify the algebraic expression $\frac{(x+2)!}{x!}$ and then solve the equation $\frac{(x+2)!}{x!} = 72$.

Solve

The pigeonhole principle

If you have more pigeons than pigeonholes, at least one pigeonhole will contain two or more pigeons. It seems obvious, but it is a very powerful tool for proofs: it lets you guarantee that something repeats, without needing to find the specific case.

Illustrative diagram of the pigeonhole principle with objects distributed into boxes, where, since there are more objects than boxes, at least one holds more than one. 4 pigeons, 3 pigeonholes: one has 2 or more
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In a chest of drawers, there are loose socks of $3$ different colours (black, white, and blue). If there is a power outage and you take out socks in the dark, what is…

In a chest of drawers, there are loose socks of $3$ different colours (black, white, and blue). If there is a power outage and you take out socks in the dark, what is the minimum number of socks you must pull out to guarantee having at least one pair of the same colour?

Solve
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In a secondary school, there are $400$ students enrolled. Justify mathematically why you can guarantee that at least two students share the same birthday. What is the…

In a secondary school, there are $400$ students enrolled.

  • Justify mathematically why you can guarantee that at least two students share the same birthday.

  • What is the minimum number of students needed at the school to guarantee that at least three of them share the same birthday? (Assume non-leap years of $365$ days).

Solve
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$5$ points are chosen at random inside a square of side length $2\text{ cm}$. Prove, using the pigeonhole principle, that there will always be at least two points whose…

$5$ points are chosen at random inside a square of side length $2\text{ cm}$. Prove, using the pigeonhole principle, that there will always be at least two points whose distance apart is less than or equal to $\sqrt{2}\text{ cm}$. (Hint: divide the square into strategic regions).

Solve

Back to the start: If a password has 7 positions and in each one you can choose from the 10 digits (0 to 9), by the multiplication principle there are $10^7 = 10\,000\,000$ possible combinations.

Summary

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