Letters that look the same
The letters of ANNA are A, N, N and A. If all four were different, they could be arranged in 4! = 4 × 3 × 2 × 1 = 24 orders.
But the two As look the same, and so do the two Ns. Swapping the two As, or the two Ns, gives a word that cannot be told apart from the one before. So far fewer than 24 different arrangements can be seen.
Listed, they are AANN, ANAN, ANNA, NAAN, NANA and NNAA: 6 arrangements.
ANNA in four places. The two As are shaded alike and so are the two Ns; swapping a pair that is shaded alike leaves the word unchanged.
Divide by the swaps
Label the letters for a moment as , , and , so that all four are different. Each arrangement you can see, such as NAAN, matches 4 of the 24 labeled orders: , , and . The As can be ordered 2! = 2 ways and the Ns 2! = 2 ways, and 2 × 2 = 4.
Every arrangement is counted 4 times, so there are 24 ÷ 4 = 6 different ones, which matches the list.
In general, n things in which one item is repeated a times, another b times, and so on, can be arranged in n! ÷ (a! × b! × …) different ways. BANANA has 6 letters with three As and two Ns: 6! ÷ (3! × 2!) = 720 ÷ 12 = 60. LEVEL has 5 letters with two Ls and two Es: 5! ÷ (2! × 2!) = 120 ÷ 4 = 30.
The 24 orders of four different letters, divided by the 2! swaps of the As and the 2! swaps of the Ns, leave 6 arrangements.
Two that must sit together
A restriction is a rule about where things may go. Four books, A, B, C and D, go on a shelf, and A and B must sit next to each other.
Tie A and B into one block. Now there are 3 items to arrange, the block, C and D, in 3! = 6 orders. Inside the block the pair can sit as AB or as BA, 2 ways. Each of the 6 orders comes with either, so there are 6 × 2 = 12 orders with A and B together.
The tied pair as one shaded block, with C and D: 3 items in 3! = 6 orders, each doubled because the block can read AB or BA.
Apart is everything else
With no rule, the four books sit in 4! = 24 orders. A and B are either together or apart, so they are apart in 24 − 12 = 12 orders.
Six friends in a row work the same way: 6! = 720 orders in all, 2 × 5! = 240 with two of them together, and 720 − 240 = 480 with those two apart.
A block of identical letters
The two ideas can meet. Arrange BANANA with its three As next to each other. Tie AAA into one block; the items are then AAA, B, N and N, four items with the N repeated, so there are 4! ÷ 2! = 24 ÷ 2 = 12 arrangements.
This time the block is not multiplied by the orders inside it. The three As look the same, so swapping them inside the block changes nothing. Two different people in a block can swap and be seen to swap; identical letters cannot.
The usual mistakes
Dividing by only one repeat. ANNA has two repeated letters, so divide 24 by 2! × 2! = 4 to get 6; dividing by 2 alone gives 12.
Adding the repeats instead of multiplying them. Every swap of the As can go with every swap of the Ns, so divide by 2! × 2!, not 2! + 2!.
Counting the original spelling again. The 6 arrangements of ANNA already include ANNA itself.
Forgetting that a tied pair of different things can swap. Six orders of the block, C and D become 12 once AB and BA are both counted.
Using every order of all the books when some must sit together. 4! = 24 includes the 12 orders that separate A and B.
Letter tiles and a photograph
In the first application below, the seven tiles of SUCCESS have three S tiles and two C tiles, and the S tiles are then tied into one block that is not multiplied by its orders.
In the second, six friends stand in a row with two of them together, and the count for apart is taken from all 6! orders.
Worked example: Letter Tiles Spelling SUCCESS, Where Repeated Letters Make Many Orders Look the Same
Question In a word game, a player has the seven tiles S, U, C, C, E, S, S. (a) How many different seven-letter arrangements of the tiles can be made? (b) In how many of these arrangements are the three S tiles next to each other?
1.If the seven tiles were all different, they could be arranged in 7! = 5040 orders.
If the seven tiles were all different, they would have 7! = 5040 orders. 2.The three S tiles can be swapped among themselves in 3! = 6 ways and the two C tiles in 2! = 2 ways, and none of these swaps changes what the word looks like.
Swapping the S tiles, 3! ways, or the C tiles, 2! ways, leaves the word looking the same. 3.(a) Each different arrangement was counted 6 × 2 = 12 times, so there are 504012 = 420 different arrangements.
(a) 50403! × 2! = 504012 = 420 different arrangements. 4.For (b), treat SSS as one tile. The tiles are then SSS, U, C, C and E: 5 tiles with the C repeated twice, so there are 5!2! = 1202 = 60 arrangements.
With SSS as one tile there are 5 tiles, and the C is repeated: 5!2! = 60. 5.(b) 60 of the 420 arrangements have the three S tiles together. Check: 60420 = 17. The three S tiles take 3 of the 7 places in 73 = 35 ways, and 5 of those are side by side: 535 = 17.
(b) 60 arrangements have the three S tiles together.
Answer: (a) 420 arrangements; (b) 60 arrangements
Common mistakes
- Dividing by 3! + 2! = 8 instead of 3! × 2! = 12. Every swap of the S tiles can be combined with every swap of the C tiles, so the numbers of swaps multiply.
- Multiplying the 60 in (b) by 3! for the orders inside the SSS block, as for two people who stand together. The three S tiles look the same, so their orders inside the block give only one arrangement.
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!.