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

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VJesus12: Legendary Member; Posts: 2276; Joined: Sat Oct 14, 2017 6:10 am; Followed by:3 members

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

by VJesus12 » Tue Mar 27, 2018 2:35 am

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

The OA is the option C.

I don't know how to solve this PS question. Is it 7! ? I need help. <i class="em em-confused"></i>

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Source: Beat The GMAT — Problem Solving | Tue Mar 27, 2018 2:35 am

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by GMATGuruNY » Tue Mar 27, 2018 4:21 am

VJesus12 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

The clause in red seems to imply that repeated letters are IDENTICAL.
Thus, the three A's are identical and the two B's are identical

Number of ways to arrange 7 letters = 7!.
But 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 positions, the arrangement doesn't change.
Here, we must divide by 3! to account for the three identical A's and by 2! to account for the two identical B's:
7!/(3!2!) = 420.

The correct answer is C.

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pm; Location: Vancouver, BC; Thanked: 5254 times; Followed by:1268 members; GMAT Score:770

by Brent@GMATPrepNow » Tue Mar 27, 2018 6:51 am

VJesus12 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

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8083; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Wed Mar 28, 2018 10:14 am

VJesus12 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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