A code is to be made by arranging 7 letters. Three of the letters used

IMO: C
We have 7 letters. 3 of A, 2 of B
So 7! / 3! * 2! = 420

AAA BB C D

Different Codes = 7! / (3! * 2!) = 420

Answer: option C

We can use the below formula for this calculation.

Permutation of n thingso of which P1 are alike of one kind , p2 are alike of 2nd kind and so on.

Number of permutation = n!/p1!*P2!

So the answer for this question will be 7!/3!*2! (As numbe rof identical A is 3 and identical B is 2)

Answer is 420. C

We need to determine the number of arrangements of:

A-A-A-B-B-C-D

Since we have 7 total letters and 3 repeated A’s and 2 repeated B’s, we can arrange the letters in the following number of ways, using the indistinguishable permutations formula:

7!/(3! x 2!) = (7 x 6 x 5 x 4)/2 = 7 x 3 x 5 x 4 = 21 x 20 = 420.

Answer: C

Hi All,

We're told that a code is to be made by arranging 7 letters: three of the letters used will be the letter A, two of the letters used will be the letter B, one of the letters used will be the letter C, and one of the letters used will be the letter D - and that there is only one way to present each letter. We're asked for the number of different codes that are possible. This question is based on a specific Permutation rule. I'm going to start with a much simpler example of the rule before applying the rule to this question.

Imagine if you had just 3 letters: two As and one B and the A's can only be written in one way (re: they're identical). How many different 3-letter codes could you form?

IF... we actually had 3 DIFFERENT letters - for example: A, B and C, then there would be (3)(2)(1) = 3! = 6 possible codes. You could also write them all out:

ABC
ACB
BAC
BCA
CAB
CBA

With two identical letters though, the number of possible codes decreases. Instead of 6, it's....

AAB
ABA
BAA

...THREE possible codes. From a math-standpoint, the way that we reduce from the number 6 is to DIVIDE by the factorial of whatever letter is duplicated. Here that would be...

3!/2! since there are two As. that gives us (3)(2)(1) / (2)(1) = 3 possible codes

Using that same rule with this prompt, we have a 7-letter code, but three identical As and two identical Bs. Thus, we divide 7! by 3! and 2! This gives us...

(7)(6)(5)(4)(3)(2)(1) / (3)(2)(1)(2)(1) = (7)(6)(5)(4) / (2)(1) = (7)(6)(5)(2) = 420 possible codes

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Hi Bunuel I didn't really understand the question but guessed my way to the correct answer by using the formula n!/p!q!. The way I understand, C. 420 is the answer to this question:

A code is to be made by arranging 7 letters. Three of the letters used will be the letter A, two of the letters used will be the letter B, one of the letters used will be the letter C, and one of the letters used will be the letter D. How many distinct codes are possible?

The last bit of the original question threw me off: "If there is only one way to present each letter..." I'm not able to even begin to understand what it means. Do questions with such wording appear on GMAT?

(Please pardon if I'm unnecessarily complicating a simple question. I had never studied permutations & combinations my life until this week. Maybe I've not grasped the concept well.)

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