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A security company can use number 1 -7 to create a 5 digit [#permalink]

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23 Feb 2010, 16:39

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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. _________________

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

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!).

Re: A security company can use number 1 -7 to create a 5 digit [#permalink]

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07 Nov 2014, 09:40

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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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