2
$\begingroup$

Attempt:

  1. If the Ns appear next to each other

I treat as if there is 1 B, 3 As, 1 N which leads to $\frac{5!}{1!3!1!}$

  1. Every N is followed by an A

I count the possible "unique characters" as NA-2, A-1, B-1 so I end up with $\frac{4!}{2!}$

  1. the two Ns must be separated

No matter how much I think about it, I'm not sure I can get an answer that doesn't lead me with a headache.

  • 0
    Why are you trying to solve in this way? – 2017-02-01
  • 0
    Possible duplicate of [how many ways can the letters in ARRANGEMENT can be arranged](http://math.stackexchange.com/questions/236471/how-many-ways-can-the-letters-in-arrangement-can-be-arranged) – 2017-02-01
  • 0
    For #3, arrangements with Ns separated = all possible arrangements - arrangements with Ns together. – 2017-02-01

1 Answers 1

3

We have simply -

$\frac {6!}{3! × 2!}$ to arrange alphabets of BANANA.

Dividing by 3! and 2! as we 3 A's and 2 N's.