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805+ (Hard)|   Combinations|      
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rsrighosh
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For a certain alarm system, each code is comprised of 5 digi 28 Oct 2021, 07:06
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VeritasKarishma

Great explanation Karishma. Thanks a lot. It's very much clear now.
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Mugdho
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For a certain alarm system, each code is comprised of 5 digi 29 Dec 2021, 06:41
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KarishmaB
VeritasPrepKarishma
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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I am still not getting why we can't write "Case 3 (aabbc): 10*10*9*9*8 = 64800 ways"

However I understand the combination approach. Can anyone plz elaborate little more the above case ?

Bunuel
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For a certain alarm system, each code is comprised of 5 digi 29 Dec 2021, 13:56
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KarishmaB and Bunuel

Can you explain where I am going wrong in the calculation for the scenario AABCD (3 distinct, one duplicate). I see how your logic works, but I am not sure why this logic is failing:

Pick first four distinct numbers:
10*9*8*7*(fifth)

Since the fifth item must come from one of the 4 already picked distinct items, there are 4 possible values. We then divide by 2! to account for the duplicate:
10*9*8*7*4 / 2 = 10*9*8*7*2 as the final answer

This differs from your number. Where am I going wrong in this logic? This logic works well for the first scenario (all distinct) but also fails in the third scenario, below:

Scenario of 1 distinct, 2 repeats (AABBC)
(10*9*8*2*1)/(2*2) =
(10*9*4)
360
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For a certain alarm system, each code is comprised of 5 digi 23 Nov 2024, 00:07
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Is there any way to do this using permutation? For Case 2, 9P3 is 504 x 10P1 = 5,040 but I don't know how to find the extra factor of 10.

KarishmaB
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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For a certain alarm system, each code is comprised of 5 digi 12 Dec 2025, 10:47
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