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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] New post 18 Oct 2010, 09:35
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

28% (01:48) correct 71% (01:01) wrong based on 14 sessions
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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Re: digit codes combination [#permalink] 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] 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] New post 18 Oct 2010, 10:40
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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] 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] 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] New post 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] New post 25 Sep 2013, 22:51
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Re: digit codes combination   [#permalink] 25 Sep 2013, 22:51
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