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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

1

This post received KUDOS

Expert's post

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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.

_________________

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

Expert's post

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.

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.

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Re: digit codes combination
[#permalink]
01 Jun 2014, 09:02