Three books on a shelf
Three different books, A, B and C, go on a shelf in a row. Any of the 3 can go first. Once one is placed, 2 are left for the second place, and then only 1 for the third.
So there are 3 × 2 × 1 = 6 orders. Listed, they are ABC, ACB, BAC, BCA, CAB and CBA.
One box for each place on the shelf. Each book placed leaves one fewer for the next box: 3, then 2, then 1.
The first book has 3 branches and each of them 2 more. The third book is whichever is left, so each tip is one complete order: 6 in all.
n factorial
The product of the whole numbers from n down to 1 is called n factorial and written n!. So 3! = 3 × 2 × 1 = 6, and 5! = 5 × 4 × 3 × 2 × 1 = 120.
n! is the number of ways to arrange n different things in a row: n choices for the first place, n − 1 for the second, and so on down to 1 for the last. Five different books go on a shelf in 120 orders.
Each factorial is the one before multiplied by n, because one more thing adds one more place at the front with n choices: 5! = 5 × 4! = 5 × 24 = 120, and 6! = 6 × 120 = 720.
Five places with 5, 4, 3, 2 and 1 choices: 5! = 120.
The factorials from 0! to 8!. Each row is the row above multiplied by its own n: 8! = 8 × 5040 = 40320.
Zero factorial
There is exactly one way to arrange nothing: leave the shelf empty. So 0! = 1.
The rule n! = n × (n − 1)! agrees. Run it backward by dividing: 3! = 4! ÷ 4 = 6, 2! = 3! ÷ 3 = 2, 1! = 2! ÷ 2 = 1, and 0! = 1! ÷ 1 = 1. Formulas for counting divide by factorials, and they rely on 0! = 1 when nothing is left over.
Canceling factorials
A quotient of factorials is worked out by canceling, not by multiplying out. 7! ÷ 5! = (7 × 6 × 5 × 4 × 3 × 2 × 1) ÷ (5 × 4 × 3 × 2 × 1) = 7 × 6 = 42, because the factors from 5 down to 1 are on the top and the bottom.
In the same way 10! ÷ 8! = 10 × 9 = 90, and 6! ÷ 3! = 6 × 5 × 4 = 120.
The usual mistakes
Giving every place n choices. Four books on a shelf give 4 × 3 × 2 × 1 = 24 orders, not 4 × 4 × 4 × 4 = 256, because each book placed cannot be placed again.
Stopping one book short. 3! = 6 arranges three books; a fourth book multiplies the count by 4, to 24.
Adding instead of multiplying. 5! is 5 × 4 × 3 × 2 × 1 = 120, not 5 + 4 + 3 + 2 + 1 = 15.
Taking 0! as 0. There is one way to arrange nothing, so 0! = 1.
Six friends in a photograph
In the application below, six friends stand in a row in 6! = 720 orders. Two of them must stand together, so they are tied into one block, which leaves 5 items to arrange in 5! = 120 orders.
Worked example: Six Friends in a Row for a Photograph, Where Two of Them Must Stand Together
Question Six friends, including Amir and Bea, stand in a row for a photograph. (a) In how many different orders can they stand if Amir and Bea must stand next to each other? (b) In how many orders are Amir and Bea not next to each other?
1.Tie Amir and Bea into one block. The block and the other 4 friends are 5 items, which stand in a row in 5! = 5 × 4 × 3 × 2 × 1 = 120 orders.
Tie Amir and Bea into one block: the block and the other 4 friends stand in 5! = 120 orders. 2.Inside the block, Amir can stand on the left or on the right: 2! = 2 orders.
Inside the block, the two can stand in 2! = 2 orders. 3.(a) Multiply the two choices: 2 × 120 = 240 orders with Amir and Bea together.
(a) 2 × 120 = 240 orders with Amir and Bea together. 4.Without any rule, the six friends stand in 6! = 720 orders.
With no rule, the six friends stand in 6! = 720 orders. 5.(b) Every order has the two either together or apart, so apart is 720 − 240 = 480 orders. Check: 240720 = 13, and of the 62 = 15 pairs of places the two can take, 5 are side by side, which is also 13.
(b) Apart is all orders minus together: 720 − 240 = 480.
Answer: (a) 240 orders; (b) 480 orders
Common mistakes
- Forgetting the order inside the block and answering 120. Amir then Bea and Bea then Amir are different photographs, so the 5! orders of the blocks are doubled.
- Counting (b) as 4! or 6! − 5!. The together case is 2 × 5! = 240, and only that number is taken from 6!.