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

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A code is to be made by arranging 7 letters. Three of the letters used  [#permalink]

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New post 27 Dec 2015, 08:53
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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. If there is only one way to present each letter, how many different codes are possible?

A. 42
B. 210
C. 420
D. 840
E. 5,040

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Re: A code is to be made by arranging 7 letters. Three of the letters used  [#permalink]

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New post 29 Dec 2015, 00:59
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Bunuel wrote:
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. If there is only one way to present each letter, how many different codes are possible?

A. 42
B. 210
C. 420
D. 840
E. 5,040



we have 7 letters out of which 3 are of one kind, 2 are of another kind..
so total ways = 7!/3!2!=420
ans C
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Re: A code is to be made by arranging 7 letters. Three of the letters used  [#permalink]

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New post 30 Dec 2015, 07:15
Bunuel wrote:
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. If there is only one way to present each letter, how many different codes are possible?

A. 42
B. 210
C. 420
D. 840
E. 5,040



IMO: C
We have 7 letters. 3 of A, 2 of B
So 7! / 3! * 2! = 420
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Re: A code is to be made by arranging 7 letters. Three of the letters used  [#permalink]

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New post 30 Dec 2015, 11:11
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Bunuel wrote:
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. If there is only one way to present each letter, how many different codes are possible?

A. 42
B. 210
C. 420
D. 840
E. 5,040


AAA BB C D

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

Answer: option C
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Re: A code is to be made by arranging 7 letters. Three of the letters used  [#permalink]

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New post 02 Jan 2016, 09:02
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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



Bunuel wrote:
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. If there is only one way to present each letter, how many different codes are possible?

A. 42
B. 210
C. 420
D. 840
E. 5,040
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Re: A code is to be made by arranging 7 letters. Three of the letters used  [#permalink]

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New post 27 Mar 2018, 07:54
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Bunuel wrote:
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. If there is only one way to present each letter, how many different codes are possible?

A. 42
B. 210
C. 420
D. 840
E. 5,040



-------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 want to arrange A, A, A, B, B, C, and D
There are 7 letters in total
There are 3 identical A's
There are 2 identical B's
So, the total number of possible arrangements = 7!/[(3!)(2!)
= (7)(6)(5)(4)(3)(2)(1)/(3)(2)(1)(2)(1)
= (7)(6)(5)(4)/(2)(1)
= (7)(6)(5)(2)
= 420

Answer: C

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Re: A code is to be made by arranging 7 letters. Three of the letters used  [#permalink]

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New post 28 Mar 2018, 10:37
Bunuel wrote:
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. If there is only one way to present each letter, how many different codes are possible?

A. 42
B. 210
C. 420
D. 840
E. 5,040


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
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Re: A code is to be made by arranging 7 letters. Three of the letters used  [#permalink]

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New post 24 Jun 2019, 16:27
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:

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Re: A code is to be made by arranging 7 letters. Three of the letters used   [#permalink] 24 Jun 2019, 16:27
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