Arrangements with conditions

Fill the hard places first

Many counting questions add a condition. The safest plan is to deal with the condition first, then fill the free places.

Objects that must be together

Six friends, Ria, Sam, Tara, Uday, Vani and Wasim, sit in a row. Ria and Sam must sit side by side. How many rows are possible?

Step 1. Glue. Treat Ria and Sam as one block. Now there are $5$ units: the block and the other four.

Step 2. Arrange the units. $5! = 120$ orders.

Step 3. Arrange inside the block. Ria and Sam can sit in $2! = 2$ orders.

Answer: $5! \times 2! = 240$ rows.

Objects that must stay apart

Now Ria and Sam must never sit side by side. Count all the rows, then take away the rows where they are together:

This is counting by complement. It is often easier to count what you do not want.

When three or more objects are glued, the complement counts rows where they are not all together. Some of those rows may still have two of them side by side.

Raise $k$ to $n$. Everyone must then stand together, which is no condition at all, so the red bar falls to $0$.

Vowels together, and vowels apart

The word DAUGHTER has $8$ different letters, with the vowels A, U and E.

• Vowels together. Glue AUE. The block and the $5$ consonants make $6$ units, so $6! \times 3! = 720 \times 6 = 4320$.
• Vowels not all together. $8! - 4320 = 40320 - 4320 = 36000$.

Places that are fixed

Sometimes the condition names the places. Arrange the letters of ORANGE so that the vowels O, A and E stand only in the odd places $1$, $3$ and $5$.

Step 1. The vowels. $3$ vowels fill $3$ odd places: $3! = 6$ ways.

Step 2. The consonants. R, N and G fill the $3$ even places: $3! = 6$ ways.

Answer: $6 \times 6 = 36$ arrangements.

Numbers from digits

Digits bring two conditions of their own.

• A number cannot start with $0$.
• An even number ends in $0$, $2$, $4$, $6$ or $8$.

How many $4$ digit numbers can be made from $0, 1, 2, 3, 4, 5$, with no digit repeated?

Step 1. The first digit. It cannot be $0$, so it has $5$ choices.

Step 2. The rest. $5$ digits are left for $3$ places: $5 \times 4 \times 3 = 60$.

Answer: $5 \times 60 = 300$ numbers.

How many of them are even? The two conditions meet at the digit $0$, so split into cases.

• Ending in 0. The other $3$ places use $5$ digits: $5 \times 4 \times 3 = 60$.
• Ending in 2 or 4. The first digit cannot be $0$ or the last digit, so it has $4$ choices. Then $4 \times 3$ for the middle. That gives $2 \times 4 \times 4 \times 3 = 96$.

Answer: $60 + 96 = 156$ even numbers.

The rank of a word

List all the arrangements of the letters of SMILE in dictionary order. In which place is SMILE itself?

The letters in alphabetical order are E, I, L, M, S. Count the words that come before SMILE.

Step 1. First letter before S. Words starting with E, I, L or M: $4 \times 4! = 96$.

Step 2. S, then a letter before M. The letters left are E, I, L and M. E, I and L come before M: $3 \times 3! = 18$.

Step 3. SM, then a letter before I. Only E comes before I: $1 \times 2! = 2$.

Step 4. SMI, then a letter before L. Only E: $1 \times 1! = 1$.

Step 5. SMIL, then E. That is SMILE itself.

Answer: $96 + 18 + 2 + 1 = 117$ words come first, so SMILE is in place $118$.

Practise with new numbers

Try it in Python

The program forms all $720$ rows of the six friends and checks the gap between Ria and Sam in each one.

Common mistakes

• Forgetting the orders inside the block. Ria and Sam can swap places within it.
• Subtracting from the wrong total. Never together is all rows minus together rows.
• Letting 0 start a number. Fill the first place before the others.
• Counting even numbers in one go. Split into cases when $0$ is one of the digits.
• Off by one in a rank. The rank is one more than the number of words before it.

Remember these

• Together: glue the objects into one block, then multiply by the orders inside it.
• Never together: all arrangements minus the together ones.
• Fixed places: fill the restricted places first, then the free ones.
• With digits, the first place cannot be $0$. Split into cases when $0$ meets another condition.
• Rank $=$ the words before it $+ 1$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Five books stand on a shelf, and two of them form a set that must stay side by side. How many orders are possible?

2. In how many of the $5! = 120$ orders are those two books apart?

3. How many $3$ digit numbers can be made from $1, 2, 3, 4$ with no digit repeated?

4. How many of those numbers are odd?

5. What is the rank of ACT among the arrangements of A, C and T in dictionary order?