How do you count arrangements where two items must stay together?

Treat the two as stuck together into a single item, and arrange that block with everything else. Then arrange the two items within the block separately, and multiply the two counts together.

True or False? "All three items must be separated" means the same as "the three items cannot all be together".

False. "Cannot all be together" still allows two of them to sit side by side, so you find it by subtracting the all-together arrangements from the total. "Completely separated" is stricter: lay out the other items first and slot these into the gaps between them.

Choosing 2 letters from A, B and C gives 6 permutations but only 3 combinations. Why?

Because AB and BA are two different arrangements but the same selection . Each combination has been counted twice over, so dividing the 6 permutations by 2 gives 3 combinations.

How do you tell whether a question is about permutations or combinations?

Ask whether the order matters : if rearranging the same objects gives a genuinely different outcome, it is a permutation. Guessing first, second and third prize is a permutation, while naming the top three scorers in any order is a combination.

When do you multiply your options, and when do you add them?

Multiply when you need one thing and another, since every choice of the first pairs with every choice of the second. Add when you need one thing or another, since those are separate cases that do not combine.

How do you handle a restriction such as "at least one yellow and no more than three reds"?

Break it into every case that satisfies the restriction, such as one yellow with three reds, then two yellows with two reds, and so on. Multiply within each case, because both conditions must hold, then add the cases together, because any one of them will do.

True or False? A question asking "how many ways" is always a combinations question.

False. "How many ways" is used for both types, so it settles nothing on its own. Look instead for whether order matters, or for the stronger keywords such as arrange against choose .

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Permutations & Combinations (Cambridge (CIE) IGCSE Additional Maths): Flashcards

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FrontPermutations
How many ways can $n$ different objects be arranged in a row?

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Cards in this collection (24)

How many ways can $n$ different objects be arranged in a row?
There are $n \times (n - 1) \times (n - 2) \times \dots \times 2 \times 1$ ways.
The count drops by one each time because every object you place is no longer available for the next position.
How many distinct arrangements are there of the letters in MATHS?
There are $5 \times 4 \times 3 \times 2 \times 1 = 120$ .
Five different letters can go first, then four remain for the second place, and so on down to one.
Define $n$ factorial.
$n!$ means $n \times (n - 1) \times (n - 2) \times \dots \times 2 \times 1$ , for any non-negative integer $n$ .
It is an operation like $+$ or $\times$ , and it is exactly the number of ways of arranging $n$ different objects.
True or False?
$0! = 1$
True.
There is exactly one way to arrange nothing at all, namely the empty arrangement, so the count is 1 rather than 0.
The factorial of a negative number, by contrast, is simply not defined.
Complete these factorial results:
$\frac{n!}{(n - 1)!} =______$
$\frac{n!}{(n - 2)!} =______$
The completed results are:
$\frac{n!}{(n - 1)!} = n$
$\frac{n!}{(n - 2)!} = n (n - 1)$
Everything from $(n - 1)$ downwards cancels, leaving only the factors at the top end.
Simplify $\frac{7!}{4!}$ by cancelling.
Writing both out in full, everything from 4 downwards is common to the top and bottom:
$\frac{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1} = 7 \times 6 \times 5$
Dividing a large factorial by a smaller one always collapses like this, so never work the two out separately.
Complete the formula for the number of permutations of $r$ objects chosen from $n$ :
${}^{n}P_{r} =______$
The completed formula is:
${}^{n}P_{r} = \frac{n!}{(n - r)!}$
A permutation counts arrangements, so the order of the $r$ chosen objects matters.
Why is arranging 3 out of 5 objects the same as $\frac{5!}{2!}$ ?
Counting directly gives $5 \times 4 \times 3$ , since there are five choices, then four, then three.
Dividing $5!$ by $2!$ cancels the $2 \times 1$ at the bottom end, leaving exactly $5 \times 4 \times 3$ .
How do you count arrangements where two items must stay together?
Treat the two as stuck together into a single item, and arrange that block with everything else.
Then arrange the two items within the block separately, and multiply the two counts together.
True or False?
"All three items must be separated" means the same as "the three items cannot all be together".
False.
"Cannot all be together" still allows two of them to sit side by side, so you find it by subtracting the all-together arrangements from the total.
"Completely separated" is stricter: lay out the other items first and slot these into the gaps between them.
Define a combination.
A combination is a selection of $r$ objects from $n$ in which the order does not matter.
That is the whole difference from a permutation, where a different order counts as a different arrangement.
Choosing 2 letters from A, B and C gives 6 permutations but only 3 combinations. Why?
Because AB and BA are two different arrangements but the same selection.
Each combination has been counted twice over, so dividing the 6 permutations by 2 gives 3 combinations.
Complete the formula for the number of combinations of $r$ objects from $n$ :
${}^{n}C_{r} =______$
The completed formula is:
${}^{n}C_{r} = \frac{n!}{(n - r)! r!}$
It is also written $(\binom{n}{r})$ and read as n choose r.
Why does the combination formula divide by $r!$ ?
Because ${}^{n}P_{r}$ counts every selection once for each way of ordering it, and there are $r!$ such orderings.
Dividing by $r!$ removes that duplication, which is why ${}^{n}C_{r} = \frac{{}^{n}P_{r}}{r!}$ .
What are ${}^{n}C_{n}$ and ${}^{n}C_{0}$ , and why?
Both equal 1.
There is only one way to choose all $n$ objects, and only one way to choose none of them.
Working these out from the formula also shows why $0!$ has to be 1.
True or False?
${}^{n}C_{r} = {}^{n}C_{n - r}$
True.
Choosing which $r$ objects to take is the same decision as choosing which $n - r$ to leave behind.
This symmetry is a useful shortcut, since ${}^{20}C_{18}$ is much quicker worked as ${}^{20}C_{2}$ .
Why is ${}^{n}C_{r}$ also called a binomial coefficient?
Because the same numbers appear as the coefficients when a two-term bracket is raised to a power.
So the quantity that counts selections is exactly the quantity that multiplies each term of a binomial expansion.
How do you tell whether a question is about permutations or combinations?
Ask whether the order matters: if rearranging the same objects gives a genuinely different outcome, it is a permutation.
Guessing first, second and third prize is a permutation, while naming the top three scorers in any order is a combination.
Fill in the type each set of keywords points to:
arrange, order, sequence suggest $______$ .
select, choose, group suggest $______$ .
The completed sentences are:
arrange, order, sequence suggest permutations.
select, choose, group suggest combinations.
Keywords are a strong hint rather than a guarantee, so check the context as well.
When do you multiply your options, and when do you add them?
Multiply when you need one thing and another, since every choice of the first pairs with every choice of the second.
Add when you need one thing or another, since those are separate cases that do not combine.
A four-digit PIN is built from ten digits. How does the count change if digits may be repeated?
With repetition allowed every position still has all ten options, giving $10 \times 10 \times 10 \times 10$ .
Without repetition the options fall away each time, giving $10 \times 9 \times 8 \times 7$ , so check this before choosing any formula.
A password is 2 letters from A to E followed by 3 digits from 0 to 9, with no repeats. How many are there?
Order matters in a password, so each part is a permutation: ${}^{5}P_{2}$ for the letters and ${}^{10}P_{3}$ for the digits.
You need letters and digits, so multiply:
$(5 \times 4) \times (10 \times 9 \times 8) = 14400$
How do you handle a restriction such as "at least one yellow and no more than three reds"?
Break it into every case that satisfies the restriction, such as one yellow with three reds, then two yellows with two reds, and so on.
Multiply within each case, because both conditions must hold, then add the cases together, because any one of them will do.
True or False?
A question asking "how many ways" is always a combinations question.
False.
"How many ways" is used for both types, so it settles nothing on its own.
Look instead for whether order matters, or for the stronger keywords such as arrange against choose.

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