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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]

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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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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

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Joined: 02 Sep 2009
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23 Feb 2010, 16:20
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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23 Feb 2010, 16:42

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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Status: No Progress without Struggle
Joined: 04 Aug 2017
Posts: 43
Location: Armenia
GPA: 3.4
Re: A security company can use number 1 -7 to create a 5 digit  [#permalink]

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28 Aug 2018, 09:28
It's a relatively easy permutation problem that includes two restrictions. The order here matters because logically the code will 76543 will be different from 76534, so we possibilities for choice are high too.
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What you feel, you attract,
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Re: A security company can use number 1 -7 to create a 5 digit &nbs [#permalink] 28 Aug 2018, 09:28
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