Jul 21 07:00 AM PDT  09:00 AM PDT Attend this webinar to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes Jul 26 08:00 AM PDT  09:00 AM PDT The Competition Continues  Game of Timers is a teambased competition based on solving GMAT questions to win epic prizes! Starting July 1st, compete to win prep materials while studying for GMAT! Registration is Open! Ends July 26th Jul 27 07:00 AM PDT  09:00 AM PDT Learn reading strategies that can help even nonvoracious reader to master GMAT RC
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 56306

A code is to be made by arranging 7 letters. Three of the letters used
[#permalink]
Show Tags
27 Dec 2015, 08:53
Question Stats:
76% (01:28) correct 24% (01:57) wrong based on 161 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.
_________________



Math Expert
Joined: 02 Aug 2009
Posts: 7763

Re: A code is to be made by arranging 7 letters. Three of the letters used
[#permalink]
Show Tags
29 Dec 2015, 00:59
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
_________________



Manager
Joined: 12 Mar 2015
Posts: 82
Concentration: Leadership, Finance
GPA: 3.9
WE: Information Technology (Computer Software)

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



CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2960
Location: India
GMAT: INSIGHT
WE: Education (Education)

Re: A code is to be made by arranging 7 letters. Three of the letters used
[#permalink]
Show Tags
30 Dec 2015, 11:11
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
_________________
Prosper!!!GMATinsightBhoopendra Singh and Dr.Sushma Jha email: info@GMATinsight.com I Call us : +919999687183 / 9891333772 Online OneonOne Skype based classes and Classroom Coaching in South and West Delhihttp://www.GMATinsight.com/testimonials.htmlACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION



Manager
Joined: 02 Jun 2015
Posts: 62
Location: United States
Concentration: Strategy, Human Resources
WE: Engineering (Manufacturing)

Re: A code is to be made by arranging 7 letters. Three of the letters used
[#permalink]
Show Tags
02 Jan 2016, 09:02
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



CEO
Joined: 12 Sep 2015
Posts: 3854
Location: Canada

Re: A code is to be made by arranging 7 letters. Three of the letters used
[#permalink]
Show Tags
27 Mar 2018, 07:54
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 Cheers, Brent
_________________
Test confidently with gmatprepnow.com



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 6967
Location: United States (CA)

Re: A code is to be made by arranging 7 letters. Three of the letters used
[#permalink]
Show Tags
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: AAABBCD 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
_________________
5star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button.



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 14603
Location: United States (CA)

Re: A code is to be made by arranging 7 letters. Three of the letters used
[#permalink]
Show Tags
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 3letter 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 mathstandpoint, 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 7letter 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
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****
Rich Cohen
CoFounder & GMAT Assassin
Special Offer: Save $75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/




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






