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A security company can use number 1 -7 to create a 5 digit

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A security company can use number 1 -7 to create a 5 digit [#permalink] New post 23 Feb 2010, 15:39
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A security company can use number 1 -7 to create a 5 digit lock combo. No numbers can be repeated and the first and last digit must be odd. How many unique lock combos can be created?

Ok, in the MGMAT guide on combinatronics they have two similar problems with & IMO they use two routes in solving them. This is frustrating me. I'm counting on the resident experts here at GMATclub to supply me with the defacto solution.
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Re: Combintonics Problem [#permalink] New post 23 Feb 2010, 15:50
Currency wrote:
Ok, in the MGMAT guide on combinatronics they have two similar problems with & IMO they use two routes in solving them. This is frustrating me. I'm counting on the resident experts here at GMATclub to supply me with the defacto solution.

Problem:

A security company can use number 1 -7 to create a 5 digit lock combo. No numbers can be repeated and the first and last digit must be odd. How many unique lock combos can be created?



Lets take it this way, first and last digit needs to be odd so that means it take 1,3,5,7

So, for first digit we have 4 options and for last digit we have 3 digits

for the rest of the 3 digits we have 5 digits to choose from so 5C3= 10 options

Total combinations = 4 * 3 * 10 = 120

Cheers
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Re: Combintonics Problem [#permalink] New post 23 Feb 2010, 16:20
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nitishmahajan wrote:
Currency wrote:
Ok, in the MGMAT guide on combinatronics they have two similar problems with & IMO they use two routes in solving them. This is frustrating me. I'm counting on the resident experts here at GMATclub to supply me with the defacto solution.

Problem:

A security company can use number 1 -7 to create a 5 digit lock combo. No numbers can be repeated and the first and last digit must be odd. How many unique lock combos can be created?



Lets take it this way, first and last digit needs to be odd so that means it take 1,3,5,7

So, for first digit we have 4 options and for last digit we have 3 digits

for the rest of the 3 digits we have 5 digits to choose from so 5C3= 10 options

Total combinations = 4 * 3 * 10 = 120

Cheers


We have 7 numbers: 1-2-3-4-5-6-7, out of which 4 are odd. Want to make 5 digit code, plus first and last digits must be odd: OXXXO.

#1:
P^2_4*P^3_5=12*60=720

P^2_4 - choosing 2 odds out of 4 odds, (note that order matters in this case: 1---3 is different from 3---1);

P^3_5 - choosing 3 numbers (for XXX) out of 5 left, (order matters: o234o is different from o342o).

OR #2:
4 choices for the first slot, 3 choices for the last slot (4 odds-1=3), 5 choices for second slot (7 numbers -2 odds), 4 choices for the third slot and 3 choices for the fourth slot: 4*3*5*4*3=720.

OR #3:
C^2_4*2!*C^3_5*3!=720, this is basically the same as the first approach, but here we choose 2 odds out of 4 and then arranging them (2!) and respectively we choose 3 other numbers from 5 left and then arranging them (3!).

Hope it's clear.
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Re: Combintonics Problem [#permalink] New post 23 Feb 2010, 16:42
720 is the answer.

They use a "slot technique":

Ie. Spaces _,_,_,_,_

Space one has 4 options (1,3,5,7) and the last space has 3 (the 3 non-chosen odds from space one)

So we're sitting at 4,_,_,_,3

So between the first and last number been chosen 2 out of the 7 possibilites have been used leaving us with

4*5*4*3*3=720 combos

Ahhhhh, my problem on the related question is that I assumed the numbers were 1-9 when it was actually 0-9.

FRIG!!!

Thanks for the help guys!!
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Round 1: 05/12/10 handling-a-grenade-thesituation-s-official-debrief-94181.html

Round 2: 07/10/10 - This time it's personal.

Re: Combintonics Problem   [#permalink] 23 Feb 2010, 16:42
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