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Difficulty:
95%
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Question Stats:
29% (02:30) correct
71%
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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
A. 30,240
B. 60,480
C. 91,440
D. 98,240
E. 101,040
19 Oct 2010, 08:32
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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.
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.
General Discussion
18 Oct 2010, 10:58
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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 ?
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 ?
