Equations and harder arrangements · Permutation & combination

General Maths gave you the tools (see counting): the multiplication rule, $n!$ , ${}^nP_r$ for arrangements and ${}^nC_r$ for selections.

{}^nP_r = \frac{n!}{(n - r)!} \qquad {}^nC_r = \frac{n!}{r!\,(n - r)!}

Arrangements and selectionsEach selection of r gives r! arrangements: ⁿPᵣ = ⁿCᵣ × r!

Further Maths uses them with an unknown $n$ , and in arrangements with conditions.

Equations in ⁿPᵣ and ⁿCᵣ

Write each symbol as a product of whole numbers, cancel the factorials, then solve. For example, ${}^nP_2 = n(n - 1)$ and ${}^nC_2 = \frac{n(n - 1)}{2}$ . Keep only a positive whole-number answer that is at least $r$ .

Worked example · WAEC 2016

WAEC 2016 · Paper 2 · Q5 (a)

If ${}^nP_5 \div {}^nC_4 = 24$ , find the value of $n$ .

Write out each symbol
- ${{}^nP_5 = \frac{n!}{(n - 5)!}}$ .
- ${{}^nC_4 = \frac{n!}{4!\,(n - 4)!}}$ .
Divide
- ${\frac{n!}{(n - 5)!} \times \frac{4!\,(n - 4)!}{n!}}$ .
- The $n!$ cancels: ${\frac{4!\,(n - 4)!}{(n - 5)!}}$ .
- ${(n - 4)! = (n - 4) \times (n - 5)!}$ , so this is ${24(n - 4)}$ .
Think first. Dividing by a fraction means multiplying by its reciprocal. What cancels?
Solve
- ${24(n - 4) = 24}$ , so ${n - 4 = 1}$ .
- So ${n = 5}$ .

More: equations and expressions in ⁿCᵣ

Objects that must stay together

When some objects must stand together, glue them into one block. Arrange the blocks and single objects as units, then multiply by the number of ways to arrange the objects inside each block:

Glue them togetherArrange the units, then arrange inside each block

Keep them togetherTick the subjects that must stay together

9units to arrange362,8809!

With no conditions, all 9 different books can go in any order: 9! = 362,880.

Worked example · WAEC 2016

WAEC 2016 · Paper 2 · Q13 (a)

Two different Mathematics books, 5 different Physics books and 3 different Chemistry books are to be arranged on a shelf. How many arrangements are possible if: (i) books on the same subject must stand together; (ii) only the Physics books must stand together?

All subjects together
- The three subject blocks can be ordered in ${3! = 6}$ ways.
- Inside: Mathematics ${2! = 2}$ , Physics ${5! = 120}$ , Chemistry ${3! = 6}$ .
- So ${6 \times 2 \times 120 \times 6 = 8640}$ arrangements.
Think first. Three blocks. How many orders for the blocks? And inside each?
Only Physics together
- Glue the 5 Physics books: the units are 1 block, 2 Mathematics and 3 Chemistry books, 6 units.
- The units: ${6! = 720}$ orders. Inside the block: ${5! = 120}$ .
- So ${720 \times 120 = 86\,400}$ arrangements.
Think first. How many units are there now?

Letters that repeat

When some of the objects are identical, such as the letters of a word with repeated letters, swapping two identical letters gives the same word. Arrange them as if they were all different, then divide by the number of ways to shuffle each set of repeats. For $n$ objects with one letter repeated $p$ times, another $q$ times, and so on:

\text{arrangements} = \frac{n!}{p!\,q!\,\cdots}

Take the tags offA, A, B: 3! orders with the A's tagged, but each word appears 2! times, so 3!/2! = 3

A letter that appears once has $1! = 1$ , so it changes nothing.

Worked example

In how many ways can the letters of the word BANANA be arranged?

Count the letters
- BANANA has 6 letters.
- A appears 3 times.
- N appears 2 times, and B once.
Think first. How many letters in all, and which of them repeat?
As if all different
- Tag the repeats: ${\text{B}, \text{A}_1, \text{N}_1, \text{A}_2, \text{N}_2, \text{A}_3}$ .
- These can be ordered in ${6! = 720}$ ways.
Think first. If every letter were different, how many orders?
Divide out the repeats
- Shuffling the three A’s gives the same word: divide by ${3! = 6}$ .
- Swapping the two N’s gives the same word: divide by ${2! = 2}$ .
- So ${\frac{720}{6 \times 2} = 60}$ arrangements.
Think first. How many times is each word counted among the 720?

More: letters that repeat

WAEC 2020 · Paper 1 · Q26In how many ways can the letters of the word MEMBER be arranged?

Numbers from given digits

Fill the most restricted place first. A number greater than 5000 restricts the first digit; an even number restricts the last digit. If both are restricted and they can clash (the same digit could be needed in both places), split into cases.

Worked example · WAEC 2019

WAEC 2019 · Paper 2 · Q1

How many even numbers greater than 5000 can be formed using the digits 1, 2, 4, 5, 6 if no digit is used more than once?

Which numbers?
- Every five-digit number from these digits is greater than 5000.
- A four-digit number must start with 5 or 6.
- Every number must end in an even digit: 2, 4 or 6.
Think first. Can a five-digit number be less than 5000? What must a four-digit one start with?
Four digits, starting with 5
- The last digit: 2, 4 or 6, which is 3 choices.
- The middle two from the 3 digits left: ${3 \times 2 = 6}$ .
- So ${3 \times 6 = 18}$ numbers.
Think first. How many choices for the last digit? Then the middle two?
Four digits, starting with 6
- The last digit: 2 or 4 (6 is used), which is 2 choices.
- The middle two: ${3 \times 2 = 6}$ .
- So ${2 \times 6 = 12}$ numbers.
Five digits
- The last digit: 3 choices. The other four digits in any order: ${4! = 24}$ .
- So ${3 \times 24 = 72}$ numbers.
- The total is ${18 + 12 + 72 = 102}$ .

More: arrangements

Your turn

WAEC 2016 · Paper 2 · Q11 (a)

(a)
If ${}^kP_2 = 72$ , find the value of $k$ .

Worked solution (try it first)

(a)

$\,{}^kP_2 = k(k - 1) = 72$ , so $k^2 - k - 72 = 0$ .
Factorise: $(k - 9)(k + 8) = 0$ .
$k$ must be a positive whole number, so $k = 9$ .

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