combinatorics

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combinatorics

by Gurpinder » Wed Jul 28, 2010 8:41 am
the security gate at a storage facility requires a five-digit lock code. if the lock code must consist only of digits from 1-7 inclusive, with no repeated digits and the first and last digits must be odd, how many lock codes are possible?


so i am using the slot method to answer this.

1. _ _ _ _ _ < 5 digits.
2. (1,3,5,7) <4 digits are odd.
3. 4 _ _ _ 3
4. (0,2,4,6) <4 digits are even so now the numbers in the middle will have more choice. 2 numbers are still available from odd numbers so....
5. 4 6 5 4 3
6. 4x6x5x4x3=1440

but the mgmat book says the answer is 720. how?
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by Brent@GMATPrepNow » Wed Jul 28, 2010 8:47 am
Gurpinder wrote:the security gate at a storage facility requires a five-digit lock code. if the lock code must consist only of digits from 1-7 inclusive, with no repeated digits and the first and last digits must be odd, how many lock codes are possible?


so i am using the slot method to answer this.

1. _ _ _ _ _ < 5 digits.
2. (1,3,5,7) <4 digits are odd.
3. 4 _ _ _ 3
4. (0,2,4,6) <4 digits are even so now the numbers in the middle will have more choice. 2 numbers are still available from odd numbers so....
5. 4 6 5 4 3
6. 4x6x5x4x3=1440

but the mgmat book says the answer is 720. how?
Your only mistake here is to include 0 as one of the numbers, when the question says to use the numbers from 1 to 7 only.

Cheers,
Brent
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by Gurpinder » Wed Jul 28, 2010 8:50 am
Brent Hanneson wrote:
Gurpinder wrote:the security gate at a storage facility requires a five-digit lock code. if the lock code must consist only of digits from 1-7 inclusive, with no repeated digits and the first and last digits must be odd, how many lock codes are possible?


so i am using the slot method to answer this.

1. _ _ _ _ _ < 5 digits.
2. (1,3,5,7) <4 digits are odd.
3. 4 _ _ _ 3
4. (0,2,4,6) <4 digits are even so now the numbers in the middle will have more choice. 2 numbers are still available from odd numbers so....
5. 4 6 5 4 3
6. 4x6x5x4x3=1440

but the mgmat book says the answer is 720. how?

Your only mistake here is to include 0 as one of the numbers, when the question says to use the numbers from 1 to 7 only.

Cheers,
Brent
omg....I should get a beating for that mistake. Thank you Brent.

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by nicolezl » Wed Jul 28, 2010 8:52 am
I agree with you up until 4_ _ _3. Then you need to choose the middle 3 numbers out of the remaining 6 digits, so the answer is:

4 X (6C3) X 3 = 12 X 6!/(2!3!) = (12 X 6 X 5 X 4)/2 = 720

Edit: Never mind, I did something wrong, since apparently I included 0 too. But I got the right answer. Can someone explain from the beginning?

Edit again: It's 4 X 5 X 4 X 3 X 3 = 720. Got it!

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by crackinggmat » Wed Jul 28, 2010 11:04 am
the first and the fifth place can be filled in 4 * 3 ways.....rest of the three places can be filled in 5*4*3 ways

hence 12 * 60 = 720