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Re: digit codes combination [#permalink]
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).
Re: digit codes combination [#permalink]
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
Re: digit codes combination [#permalink]
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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Re: digit codes combination [#permalink]
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
For a certain alarm system, each code is comprised of 5 digi [#permalink]
01 Jun 2014, 08:47
Expert's post
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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.
Re: digit codes combination [#permalink]
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
Re: For a certain alarm system, each code is comprised of 5 digi [#permalink]
11 Aug 2015, 03:32
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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).
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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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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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