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Re: Which of the following leads to the correct mathematical solution for [#permalink]
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Bunuel wrote:

Tough and Tricky questions: Combinations.



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! × 2!)
D) 6!/(3! + 2!)
E) 6!/(3! × 2!)

Kudos for a correct solution.

Source: Chili Hot GMAT


The correct answer is E.
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Which of the following leads to the correct mathematical solution for [#permalink]
Hi -
Can someone please explain why this is a combinatorics problem instead of a permutation? I know permutation is concerned with order and combinatorics is not. When we rearrange letters in a word, aren't we concerned with the order of the letters?

Thanks in advance!
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Re: Which of the following leads to the correct mathematical solution for [#permalink]
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lawiniecke wrote:
Hi -
Can someone please explain why this is a combinatorics problem instead of a permutation? I know permutation is concerned with order and combinatorics is not. When we rearrange letters in a word, aren't we concerned with the order of the letters?

Thanks in advance!


THEORY:

Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is:

\(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\).

For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \(\frac{9!}{4!3!2!}\).

For more check the links below:

Combinatorics Made Easy!

Theory on Combinations

DS questions on Combinations
PS questions on Combinations

Tough and tricky questions on Combinations
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Re: Which of the following leads to the correct mathematical solution for [#permalink]
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Bunuel wrote:

Tough and Tricky questions: Combinations.



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! × 2!)
D) 6!/(3! + 2!)
E) 6!/(3! × 2!)

Kudos for a correct solution.

Source: Chili Hot GMAT


------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
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Re: Which of the following leads to the correct mathematical solution for [#permalink]
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Re: Which of the following leads to the correct mathematical solution for [#permalink]
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