Which of the following leads to the correct mathematical solution for

Question

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

arunspanda · Answer

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!)

Number of letters in word 'BANANA' = 6.
The letters 'A' and 'N' appear 3 times and 2 times respectively in the word 'BANANA'.
Therefore the mathematical solution for number of ways that the letters of the word BANANA can be arranged to create six-letter code
= 6!/(3!*2!)

Answer:  E

JJo · Answer

BANANA - six letters can be arranged in 6! ways
since 'A' repeats 3 times and 'N' repeats 2 times, we need to divide the 6! ways by 3! and 2! to adjust repetition.

6!/(3!*2!)

Ans. E) 6!/(3! × 2!)

Bunuel · Answer

The correct answer is E.

law258 · Answer

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!

Bunuel · Answer

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

BrentGMATPrepNow · Answer

------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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Which of the following leads to the correct mathematical solution for

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