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# For a certain alarm system, each code is comprised of 5 digi

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For a certain alarm system, each code is comprised of 5 digi [#permalink]

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18 Oct 2010, 09:35
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For a certain alarm system, each code is comprised of 5 digits, but no digit can be used more than twice. How many codes can be made?

A. 30,240
B. 60,480
C. 91,440
D. 98,240
E. 101,040
[Reveal] Spoiler: OA

Last edited by Bunuel on 25 Sep 2013, 23:58, edited 1 time in total.
Renamed the topic and edited the question.
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18 Oct 2010, 09:58
3 Possibilities

a. All 5 Distinct
No. of orderings = 10C5 x 5!

b. 3 distinct, 2 of one kind
No. of orderings = 10C3 X 7C1 X (5!/2!)

c. 2 of type1, 2 of type2, and a remaining solo
No. of orderings = 10C2 X 8C1 X (5!/2!.2!)

Sum up a,b,c to get the required answer.

Is there a less tedious way to do this ?
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18 Oct 2010, 10:27
hi abhishekj2512,

I have already tried this way and although it seems right, it doesn't give the right answer.
See below:

a. All 5 Distinct = 10C5 x 5! = 30240

b. 3 distinct, 2 of one kind = 10C3 X 7C1 X (5!/2!) = 8400

c. 2 of type1, 2 of type2, and a remaining solo = 10800

10800+8400+30240 = 49400

Another way that I considered was to calculate the complementary set:
10^5 - all possible codes, and deduct the number of ways to comprise a code with a digit that repeats 3 times, 4 times and 5 times. (It didn't work as well).
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18 Oct 2010, 10:40
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b. 3 distinct, 2 of one kind = 10C3 X 7C1 X (5!/2!) = 8400

Seems like some error at your end, this comes out as 50,400
Using this, answer does come out correctly.

In this case, the forward and backward approach would involve same number of calculations.
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18 Oct 2010, 11:35
you are right. I miscalculated the multiple factorial.

Thanks.
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19 Oct 2010, 04:46
Sorry but I don't understand your answer. According to the question, no digit can be used more than twice. Then shall the answer be 10C5 = 30,240???
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19 Oct 2010, 07:32
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phamduyha: The question says 'no digit should be used more than twice' which means that a digit can be used at most 2 times. So you have to take 3 cases:

Case 1: All digits distinct
Select 5 digits out of 10 in 10C5 ways and arrange them in 5! ways which gives us 10C5x5! = 30240

Case 2: Two digits are same, other 3 are dictinct e.g. 45722
Select a digit which is to be repeated in 10C1 ways, select other 3 digits in 9C3 ways and arrange them in 5!/2! ways to get 10C1 x 9C3 x 5!/2! = 50400 ways

Case 3: Two digits repeated and one other digit e.g. 33448
Select 2 digits to be repeated in 10C2 ways, select the single digit in 8C1 ways and arrange them all in 5!/2!.2! ways to get 10C2 x 8C1 x 5!/2!.2! = 10800

Sum of all = 30240 + 50400 + 10800 = 91440 ways

Generally, actual GMAT questions will not be calculation intensive since you are required to solve them in under 2 mins.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews GMAT Club Legend Joined: 09 Sep 2013 Posts: 13489 Followers: 576 Kudos [?]: 163 [1] , given: 0 Re: digit codes combination [#permalink] ### Show Tags 25 Sep 2013, 22:51 1 This post received KUDOS Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Manager Joined: 13 Oct 2013 Posts: 137 Concentration: Strategy, Entrepreneurship Followers: 2 Kudos [?]: 37 [0], given: 125 Re: digit codes combination [#permalink] ### Show Tags 01 Jun 2014, 08:21 can someone help me here? I do not understand why we are multiplying 5! here? Select 5 digits out of 10 in 10C5 ways and arrange them in 5! ways which gives us 10C5x5! = 30240 i thought selecting 5 digits out of 10 would be just 10C5 .Am i missing anything? VeritasPrepKarishma wrote: phamduyha: The question says 'no digit should be used more than twice' which means that a digit can be used at most 2 times. So you have to take 3 cases: Case 1: All digits distinct Select 5 digits out of 10 in 10C5 ways and arrange them in 5! ways which gives us 10C5x5! = 30240 Case 2: Two digits are same, other 3 are dictinct e.g. 45722 Select a digit which is to be repeated in 10C1 ways, select other 3 digits in 9C3 ways and arrange them in 5!/2! ways to get 10C1 x 9C3 x 5!/2! = 50400 ways Case 3: Two digits repeated and one other digit e.g. 33448 Select 2 digits to be repeated in 10C2 ways, select the single digit in 8C1 ways and arrange them all in 5!/2!.2! ways to get 10C2 x 8C1 x 5!/2!.2! = 10800 Sum of all = 30240 + 50400 + 10800 = 91440 ways Generally, actual GMAT questions will not be calculation intensive since you are required to solve them in under 2 mins. _________________ --------------------------------------------------------------------------------------------- Kindly press +1 Kudos if my post helped you in any way Math Expert Joined: 02 Sep 2009 Posts: 36590 Followers: 7092 Kudos [?]: 93367 [0], given: 10557 For a certain alarm system, each code is comprised of 5 digi [#permalink] ### Show Tags 01 Jun 2014, 08:47 Expert's post 1 This post was BOOKMARKED sunita123 wrote: can someone help me here? I do not understand why we are multiplying 5! here? Select 5 digits out of 10 in 10C5 ways and arrange them in 5! ways which gives us 10C5x5! = 30240 i thought selecting 5 digits out of 10 would be just 10C5 .Am i missing anything? VeritasPrepKarishma wrote: phamduyha: The question says 'no digit should be used more than twice' which means that a digit can be used at most 2 times. So you have to take 3 cases: Case 1: All digits distinct Select 5 digits out of 10 in 10C5 ways and arrange them in 5! ways which gives us 10C5x5! = 30240 Case 2: Two digits are same, other 3 are dictinct e.g. 45722 Select a digit which is to be repeated in 10C1 ways, select other 3 digits in 9C3 ways and arrange them in 5!/2! ways to get 10C1 x 9C3 x 5!/2! = 50400 ways Case 3: Two digits repeated and one other digit e.g. 33448 Select 2 digits to be repeated in 10C2 ways, select the single digit in 8C1 ways and arrange them all in 5!/2!.2! ways to get 10C2 x 8C1 x 5!/2!.2! = 10800 Sum of all = 30240 + 50400 + 10800 = 91440 ways Generally, actual GMAT questions will not be calculation intensive since you are required to solve them in under 2 mins. Yes, the number of way to select 5 digits out of 10 is 10C5. But these 5 digits can be arranged in 5!=120 ways, each of which gives different code. Hence the number of codes for this case is 10C5*5!. One could also do 10P5 (which is basically the same as 10C5*5!): selecting 5 digits out of 10, when the order of the selection matters. Hope it's clear. _________________ Manager Joined: 13 Oct 2013 Posts: 137 Concentration: Strategy, Entrepreneurship Followers: 2 Kudos [?]: 37 [0], given: 125 Re: digit codes combination [#permalink] ### Show Tags 01 Jun 2014, 09:02 oh yess:). Thanks Bunuel. Bunuel wrote: sunita123 wrote: can someone help me here? I do not understand why we are multiplying 5! here? Select 5 digits out of 10 in 10C5 ways and arrange them in 5! ways which gives us 10C5x5! = 30240 i thought selecting 5 digits out of 10 would be just 10C5 .Am i missing anything? VeritasPrepKarishma wrote: phamduyha: The question says 'no digit should be used more than twice' which means that a digit can be used at most 2 times. So you have to take 3 cases: Case 1: All digits distinct Select 5 digits out of 10 in 10C5 ways and arrange them in 5! ways which gives us 10C5x5! = 30240 Case 2: Two digits are same, other 3 are dictinct e.g. 45722 Select a digit which is to be repeated in 10C1 ways, select other 3 digits in 9C3 ways and arrange them in 5!/2! ways to get 10C1 x 9C3 x 5!/2! = 50400 ways Case 3: Two digits repeated and one other digit e.g. 33448 Select 2 digits to be repeated in 10C2 ways, select the single digit in 8C1 ways and arrange them all in 5!/2!.2! ways to get 10C2 x 8C1 x 5!/2!.2! = 10800 Sum of all = 30240 + 50400 + 10800 = 91440 ways Generally, actual GMAT questions will not be calculation intensive since you are required to solve them in under 2 mins. Yes, the number of way to select 5 digits out of 10 is 10C5. But these 5 digits can be arranged in 5!=120 ways, each of which gives different code. Hecen the number of codes for this case is 10C5*5!. One could also do 10P5 (which is basically the same as 10C5*5!): selecting 5 digits out of 10, when the order of the selection matters. Hope it's clear. _________________ --------------------------------------------------------------------------------------------- Kindly press +1 Kudos if my post helped you in any way GMAT Club Legend Joined: 09 Sep 2013 Posts: 13489 Followers: 576 Kudos [?]: 163 [0], given: 0 Re: For a certain alarm system, each code is comprised of 5 digi [#permalink] ### Show Tags 11 Aug 2015, 03:32 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7125 Location: Pune, India Followers: 2137 Kudos [?]: 13682 [1] , given: 222 Re: For a certain alarm system, each code is comprised of 5 digi [#permalink] ### Show Tags 23 Sep 2015, 23:25 1 This post received KUDOS Expert's post VeritasPrepKarishma wrote: phamduyha: The question says 'no digit should be used more than twice' which means that a digit can be used at most 2 times. So you have to take 3 cases: Case 1: All digits distinct Select 5 digits out of 10 in 10C5 ways and arrange them in 5! ways which gives us 10C5x5! = 30240 Case 2: Two digits are same, other 3 are dictinct e.g. 45722 Select a digit which is to be repeated in 10C1 ways, select other 3 digits in 9C3 ways and arrange them in 5!/2! ways to get 10C1 x 9C3 x 5!/2! = 50400 ways Case 3: Two digits repeated and one other digit e.g. 33448 Select 2 digits to be repeated in 10C2 ways, select the single digit in 8C1 ways and arrange them all in 5!/2!.2! ways to get 10C2 x 8C1 x 5!/2!.2! = 10800 Sum of all = 30240 + 50400 + 10800 = 91440 ways Generally, actual GMAT questions will not be calculation intensive since you are required to solve them in under 2 mins. Responding to a pm: Quote: I did it in 3 cases just like you. Case1 (abcde): 10*9*8*7*6 = 30240 ways Case 2 (aabcd): 10*10*9*8*7 = 50400 ways Case 3 (aabbc): 10*10*9*9*8 = 64800 ways (Shouldn't this case has more ways than 2 cases above?) Total = 30240 + 50400 + 64800 = 145400 ways As you can see, your case 3 has a problem. There are 10 ways of writing the leftmost digit - correct Then there are 10 ways of writing the next digit - correct For the next digit, there could be 9 ways or there could be 10 ways depending on what was chosen previously: e.g 22 __ __ __ - here there are 9 ways of choosing the next digit. 23 __ __ __ - here there are 10 ways of choosing the next digit. Both 2 and 3 can be repeated as opposed to case 2 where only one digit can be repeated. Hence, don't do the question this way. Choose the repeated digits and then arrange as I have done above. Note that there will be far fewer cases here because 2 digits will be repeated so fewer different digits will be there. So the number of arrangements will be fewer. Now think, why does case 2 work but case 3 does not. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: For a certain alarm system, each code is comprised of 5 digi   [#permalink] 23 Sep 2015, 23:25
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