How many different ways can the letters “different” be arranged?
1 Answer
Explanation:
If all 9 letters were unique, then we'd have:
9 choices for the 1st letter
8 choices for the 2nd letter
...
2 choices for the 8th letter
1 choice for the 9th letter
And the number of arrangements is the product of all these choices, which is 9 factorial:
#9xx8xx7xx...xx2xx1-= 9! = "362,880"#
But, not all of the letters are unique. There are 2 f's, and there are 2 e's. Of all the 9! arrangements above, some of them are essentially the same, just with the f's (and/or e's) swapped around. For example,
#"DIFF"color(orange)"E""R"color(blue)"E""NT"# and
#"DIFF"color(blue)"E""R"color(orange)"E""NT"#
would both be counted in the 9!, but they're both the same word. We need to remove all the "double-countings".
This is done by dividing 9! by the number of ways to shuffle each copied letter. Since there are two F's, they can be shuffled 2! ways, and so can the two E's:
#(9!)/(2!xx2!) " "color(gray)("9 letters"/"(2 F's)(2 E's)")#
#="362,880"/4#
#=" 90,720"#
Another example: How many ways can the letters in BANANA be arranged?
There are 6 letters total, with 3 A's and 2 N's. Thus, the number of unique arrangements (or "words") is:
#(6!)/(3!xx2!)=(6*5*4*3*2*1)/[(3*2*1)(2*1)]=60.#
