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!)
The OA is E.
Please, can someone explain why this is a combinatorics problem instead of a permutation? I know permutation is concerned with order and combinatorics is not. When we rearranged letters in a word, aren't we concerned with the order of the letters?
Which of the following leads to the correct mathematical
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This problem does indeed require that we count the number of possible arrangements.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!)
BANANA = 6 letters.
The number of ways to arrange 6 distinct elements = 6!.
But the elements here are not all distinct.
There are 3 identical A's and 2 identical N's.
When an arrangement includes identical elements, we must divide by the number of ways each set of identical elements can be arranged.
The reason:
When the identical elements swap places, the arrangement does not change, reducing the total number of unique arrangements.
Here, we must divide by 3! to account for the 3 identical A's and by 2! to account for the 2 identical N's:
6!/(3!2!).
The correct answer is E.
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------ASIDE-----------------------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!)
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
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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!).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!)
Answer: E
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