When some items look identical, 11! counts the same word many times over.
How many arrangements
eleven letters, but not all different
Eleven letters give 11! = 39,916,800 — but M, A and T each appear twice, so divide by 2! three times: 39,916,800 ÷ 8 = 4,989,600. D forgets to divide, counting a swap of the two M's as a different word when it spells the same thing.
11 letters in total — now find out which ones show up twice.
| Letter | How many | Repeated? |
|---|---|---|
| M | 2 | yes |
| A | 2 | yes |
| T | 2 | yes |
| H, E, I, C, S | 1 each | no |
2 + 2 + 2 + 5 = 11, so nothing was missed.
Always add your tallies back up as a safety check.
Swap the two M’s and the word does not change.
first M in front
second M in front
Identical letters make identical words.
so every word was counted 2! times
once for each way of ordering the identical pair
distinct = n!2! · 2! · 2!
one factorial on the bottom for each repeated letter
Letters appearing once contribute 1!, which changes nothing.
arrangements = 11!2! · 2! · 2!
11! = 39 916 800
2 · 2 · 2 = 8
Each 2! is just 2, so the bottom stays small.
The bottom is 2! · 2! · 2! = 8, not 2 + 2 + 2 = 6. Each repeated letter multiplies the overcount.
11! = 39 916 800
the eleven-letter count
bottom = 2 · 2 · 2 = 8
three repeated letters
answer = 39 916 800 ÷ 8
now divide
= 4 989 600
the distinct words
Without dividing you would have said 39 916 800 — exactly eight times too many.
Take a word small enough to list by hand.
3 letters, D twice
formula: 3!2! = 62 = 3
ADD, DAD, DDA
Three, exactly as the formula said.
Plain 3! would have given 6, counting each word twice — the same mistake, just easier to see.
BALLOON: B, A, L × 2, O × 2, N
7 letters
fix one L in front
left to arrange: B, A, L, O, O, N
6!2! = 7202 = 360
only the O’s still repeat
After an L goes in front, only one L is left among the six, so its 2! leaves the bottom. Keeping 2! · 2! gives 180.
glue NNN into one block
block, C, A, O
4 different things: 4! = 24
nothing repeats any more
inside the block NNN can only be NNN
so no extra factor
Compare all arrangements of CANNON: 6! ÷ 3! = 120. Keeping the N’s together leaves 24 of them.
Identical letters inside a block cannot be told apart, so the block has only one inside order. 4! × 3! = 144 is more than the 120 arrangements that exist.
Not together. In how many arrangements of the letters of NOODLE are the two O’s not next to each other?
all: 6! ÷ 2! = 360
the two O’s repeat
together: block OO, N, D, L, E → 5! = 120
nothing repeats now
360 − 120 = 240
everything minus together
What is different: “apart” is counted through its opposite. The total keeps its 2! on the bottom; the glued count does not, because the block has only one inside order.
Repeated digits and a 0. How many 6-digit numbers use all of the digits 0, 0, 1, 1, 1, 2?
all orders: 6! ÷ (2! · 3!) = 720 ÷ 12 = 60
two 0’s, three 1’s
0 in front: arrange 0, 1, 1, 1, 2 → 5! ÷ 3! = 20
one 0 is used up, so its 2! goes
60 − 20 = 40
numbers cannot start with 0
What is different: digits follow the same rule as letters, plus the leading-zero rule. Remove the orders that start with 0, and recount the repeats of what is left.
60 includes strings such as 011120, which is only a 5-digit number. A question about numbers always asks you to remove the leading zeros.
Only some of the letters. How many 3-letter arrangements can be made from the letters of LEVEL (L, L, E, E, V)?
all different (L, E, V): 3! = 6
case 1
two L’s + E or V: 2 × 3!2! = 6
case 2: LLE, LEL, ELL, LLV, …
two E’s + L or V: 2 × 3!2! = 6
case 3
6 + 6 + 6 = 18
separate cases add
What is different: the formula n! ÷ (repeats) needs all the letters. With only some, split by which letters are chosen, count each case with the formula, then add.
4 steps right and 2 up
every shortest path uses exactly these 6 steps
a path = an arrangement of R R R R U U
for example RRURUR
6!4! · 2! = 72048 = 15
four R’s alike, two U’s alike
720 treats every step as different, and 720 ÷ 4! = 30 still counts each path twice. Both repeated letters go on the bottom.
Expect a twist on the plain word: a letter fixed at the front or at both ends, identical letters kept together, a word with three repeated letters, or a grid path in disguise. Tally the letters, place the fixed or glued ones first, then divide by one factorial for each letter that still repeats. Ten-second check: a restricted count must be smaller than the total for the whole word.
No repeated letter was forgotten, so the bottom is complete.
How many distinct arrangements of
10! = 3,628,800, then divide by 3! for the three S's, 3! for the three T's and 2! for the two I's: 3,628,800 ÷ 72 = 50,400. A ignores the repeats entirely. Count each repeated letter's factorial and divide by every one.
Hint: two letters appear three times here, not twice. A letter used three times contributes 3!, which is 6.
