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Statistics and Probability - Counting – Permutations – Combinations

Grade 11IB_AA

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Fundamental Counting Principle: If there are mm ways to do one thing and nn ways to do another, then there are m×nm \times n ways to do both.

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Factorial Notation: For a positive integer nn, n!=n×(n−1)×(n−2)×⋯×1n! = n \times (n-1) \times (n-2) \times \dots \times 1. Note that 0!=10! = 1.

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Permutations (nPr^n P_r): An arrangement of rr objects from a set of nn distinct objects where the order of arrangement matters.

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Combinations (nCr^n C_r): A selection of rr objects from a set of nn distinct objects where the order of selection does NOT matter.

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Arrangements with Identical Items: If there are nn objects where pp are of one type, qq are of another, etc., the number of distinct permutations is n!p!q!...\frac{n!}{p!q!...}.

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Restrictions: When certain items must be together, treat them as a single 'block'. If certain items must be separate, arrange the others first and place the restricted items in the gaps.

📐Formulae

n!=n(n−1)(n−2)…(3)(2)(1)n! = n(n-1)(n-2)\dots(3)(2)(1)

nPr=n!(n−r)!^n P_r = \frac{n!}{(n-r)!}

nCr=(nr)=n!r!(n−r)!^n C_r = \binom{n}{r} = \frac{n!}{r!(n-r)!}

Number of arrangements of n objects in a circle=(n−1)!\text{Number of arrangements of } n \text{ objects in a circle} = (n-1)!

(nr)=(nn−r)\binom{n}{r} = \binom{n}{n-r}

💡Examples

Problem 1:

How many different 4-letter 'words' can be formed using the letters of the word EQUATIONEQUATION if each letter can be used only once?

Solution:

The word EQUATIONEQUATION has 88 distinct letters. We need to arrange 44 out of these 88. This is a permutation problem since the order of letters creates different words. 8P4=8!(8−4)!=8!4!=8×7×6×5=1680^8 P_4 = \frac{8!}{(8-4)!} = \frac{8!}{4!} = 8 \times 7 \times 6 \times 5 = 1680

Explanation:

Since the order of the letters matters in a word, we use the permutation formula nPr^n P_r with n=8n=8 and r=4r=4.

Problem 2:

A committee of 55 people is to be chosen from a group of 77 men and 55 women. How many ways can the committee be formed if it must contain exactly 33 men?

Solution:

We need to choose 33 men from 77 and 22 women (to make a total of 55) from 55. Number of ways to choose men: (73)=7!3!4!=35\binom{7}{3} = \frac{7!}{3!4!} = 35 Number of ways to choose women: (52)=5!2!3!=10\binom{5}{2} = \frac{5!}{2!3!} = 10 Total ways = 35×10=35035 \times 10 = 350

Explanation:

We use combinations (nr)\binom{n}{r} because the order in which committee members are chosen does not matter. We then multiply the independent choices for men and women.

Problem 3:

Find the number of ways to arrange the letters in the word BANANABANANA.

Solution:

The word BANANABANANA has 66 letters in total. Letters: 1×B1 \times B, 3×A3 \times A, 2×N2 \times N. Total arrangements = 6!1!3!2!\frac{6!}{1!3!2!} 7201×6×2=72012=60\frac{720}{1 \times 6 \times 2} = \frac{720}{12} = 60

Explanation:

This is a permutation problem with identical items. We divide the total permutations (6!6!) by the factorials of the frequencies of the repeated letters (AA and NN).

Problem 4:

A student calculates the difference between 10!10! and 9!9! manually. Show the calculation.

Solution:

10!=362880010! = 3628800 9!=3628809! = 362880 3628800−3628803265920\begin{array}{r} 3628800 \\ - 362880 \\ \hline 3265920 \end{array}

Explanation:

The student subtracts the value of 9!9! from 10!10! using vertical subtraction.