Why should you deal with a restricted position first when counting arrangements?

Because the restriction limits which digits can go there, and fixing that position first tells you how many options are left for the rest. Leave it until last and the number of options for it depends on what you already used, so a single product no longer works.

GCSEFurther MathsAQAFlashcards1. NumberCounting Principles

Counting Principles (AQA GCSE Further Maths): Flashcards

Exam code: 8365

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FrontCounting Principles
Complete the two counting rules by filling in the missing operation:
AND means $______$
OR means $______$

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Cards in this collection (6)

Complete the two counting rules by filling in the missing operation:
AND means $______$
OR means $______$
The completed rules are:
AND means $\times$
OR means $+$
Choosing a pen and a pencil from 4 pens and 5 pencils gives $4 \times 5$ options; choosing a pen or a pencil gives $4 + 5$ .
In how many different orders can $n$ different objects be arranged in a row?
There are $n \times (n - 1) \times (n - 2) \times \dots \times 2 \times 1$ orders.
Any of the $n$ objects can go first, which leaves $n - 1$ for the second position, and so on until one object remains.
True or False?
The product rule for counting can be applied to more than two choices at once.
True.
The rule extends to any number of choices, with one factor in the product for each choice made.
A 4-letter code from 26 letters, where letters may repeat, gives $26 \times 26 \times 26 \times 26 = 26^{4}$ .
How many 4-digit PIN codes can be made from the digits 0 to 9 if digits may be repeated, and how many if they may not?
With repeats, every position has 10 options: $10 \times 10 \times 10 \times 10 = 10000$ .
Without repeats, each digit used removes an option: $10 \times 9 \times 8 \times 7 = 5040$ .
Why should you deal with a restricted position first when counting arrangements?
Because the restriction limits which digits can go there, and fixing that position first tells you how many options are left for the rest.
Leave it until last and the number of options for it depends on what you already used, so a single product no longer works.
How many 5-digit odd numbers can be made from the digits 1, 3, 4, 6 and 8, using each digit once?
There are 48.
Deal with the restriction first: to be odd the number must end in 1 or 3, so there are 2 choices for the last digit.
The other four digits fill the remaining four positions in $4 \times 3 \times 2 \times 1 = 24$ ways, and $2 \times 24 = 48$ .

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