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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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New post Updated on: 25 Sep 2013, 23:58
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A
B
C
D
E

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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

Originally posted by eladshus on 18 Oct 2010, 09:35.
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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Re: digit codes combination  [#permalink]

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New post 19 Oct 2010, 07:32
6
5
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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Re: digit codes combination  [#permalink]

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New post 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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Re: digit codes combination  [#permalink]

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New post 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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Re: digit codes combination  [#permalink]

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New post 18 Oct 2010, 10:40
1
Quote:
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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Re: digit codes combination  [#permalink]

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New post 18 Oct 2010, 11:35
you are right. I miscalculated the multiple factorial.

Thanks.
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Re: digit codes combination  [#permalink]

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New post 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???
Please explain. Thanks
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Re: digit codes combination  [#permalink]

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New post 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.

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

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New post 01 Jun 2014, 08:47
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.
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Re: digit codes combination  [#permalink]

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New post 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.

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

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New post 23 Sep 2015, 23:25
1
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.
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Re: For a certain alarm system, each code is comprised of 5 digi  [#permalink]

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New post 09 Feb 2017, 04:37
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.


Hi, could you please specify the logic for 2nd and 3rd cases.
2nd case:
I do not grasp the logic why you're counting 10C1 only once as long as we have assume that 2 digits are the same? And what is the rationale for using 5!/2! to arrange, why do we have a 2! in denominator? I'd rather used 3! to arrange remaining 3 distinct digits.
3d case:
Actually the same questions as above. I'd rather used 10C2 for the first pair of the same digits, then 8C2 for second pair of the same digits, and then 6C1 for the last one then arrange in 5!.
Would appreciate for explanation.
P.S. I'm not stable at computations. Some problems are solved smoothly however others like this one makes me stuck for many minutes...
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Re: For a certain alarm system, each code is comprised of 5 digi  [#permalink]

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New post 09 Feb 2017, 21:53
1
Alexey1989x wrote:
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.


Hi, could you please specify the logic for 2nd and 3rd cases.
2nd case:
I do not grasp the logic why you're counting 10C1 only once as long as we have assume that 2 digits are the same? And what is the rationale for using 5!/2! to arrange, why do we have a 2! in denominator? I'd rather used 3! to arrange remaining 3 distinct digits.
3d case:
Actually the same questions as above. I'd rather used 10C2 for the first pair of the same digits, then 8C2 for second pair of the same digits, and then 6C1 for the last one then arrange in 5!.
Would appreciate for explanation.
P.S. I'm not stable at computations. Some problems are solved smoothly however others like this one makes me stuck for many minutes...


You select one digit which has to be repeated. That digit will be written twice. So you just need to select one such digit. You can do that in 10C1 ways. The one selection could be 0 or 1 or 2 ...etc. This you will write twice.

Now you have place for 3 other digits. These you select in 9C3 ways.

How do you arrange something like 11234? Note that 12314, 21341 etc are all valid arrangements.
How do you arrange n things in which r are the same? You use n!/r!.

To know why, check out this video: https://www.youtube.com/watch?v=1zhltihi5VU
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Re: For a certain alarm system, each code is comprised of 5 digi  [#permalink]

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New post 21 Jan 2019, 12:46
VeritasKarishma

Hi! I understand your solution however I solved the problem focusing on the available slots instead of available numbers. I managed to get the same number as you for the first two cases... however, case three I ended up with a different number although the equation we have is equivalent. Here's what I have:

Case one: ABCDE
number of different combinations 5 numbers can make out of 10 numbers
10P5=30240
Case two: AABCD
(# of different combinations 4 numbers can make out of 10 numbers) * (#of positions that one number of each code(A) that is selected to be repeated can take within the 5 slots)
10P4 * 5C2 = 50400

Case three: AABBC
(number of different combinations 3 numbers can make out of 10 numbers) * (#of positions that one number of each code(A) that is selected to be repeated can take within the 5 slots) * (#of positions that another number(B) of each code that is selected to be repeated can take within the remaining three slots)
10P3 * 5C2 * 3C2 = (10*9*8)*[5!/(3!2!)]*(3!/2!)=21600
=> what I have broken down is essentially the same equation as you have for case three, however I got 21600 for the answer.

Please shed some light on this!
Thank you!
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