AAPL wrote:Which of the following leads to the correct mathematical solution for the number of ways that the letters of the word BANANA could be arranged to create a six-letter code?

A. 6!
B. 6! - (3! + 2!)
C. 6! - (3! x 2!)
D. 6! / (3! + 2!)
E. 6! / (3! x 2!)

------ASIDE-----------------------
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:

If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]

----------------ONTO THE QUESTION-------------------------------
We have the word BANANA:
There are 6 letters in total
There are 3 identical A's
There are 2 identical N's
So, the total number of possible arrangements = 6!/[(3!)(2!)]

Answer: E

Cheers,
Brent

AAPL wrote:Which of the following leads to the correct mathematical solution for the number of ways that the letters of the word BANANA could be arranged to create a six-letter code?

A. 6!
B. 6! - (3! + 2!)
C. 6! - (3! x 2!)
D. 6! / (3! + 2!)
E. 6! / (3! x 2!)

If there were no repeated letters, the number of rearrangements would be 6! However, the letter "A" is repeated 3 times, and the letter "N" is repeated 2 times. We thus must use the formula for indistinguishable permutations. Therefore, the number of ways to arrange the letters in BANANA is 6!/(3! x 2!).

Answer: E