Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

2

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

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. _________________

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. _________________

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.

_________________

<a href="http://gmatclub.com/forum/viewforumtags.php?sid=ea02c74d77a9ca83f3d24612723e4c68"> GMAT Club Question Bank </a>

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.

_________________

<a href="http://gmatclub.com/forum/viewforumtags.php?sid=ea02c74d77a9ca83f3d24612723e4c68"> GMAT Club Question Bank </a>

gmatclubot

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

1. My favorite football team to infinity and beyond, the Dallas Cowboys, currently have the best record in the NFL and I’m literally riding on cloud 9 because...

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...