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A door can be opened only with a security code that consists

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Director
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A door can be opened only with a security code that consists [#permalink] New post 14 Nov 2005, 18:53
00:00
A
B
C
D
E

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A door can be opened only with a security code that consists of five
buttons: 1, 2, 3, 4, 5. A code consists of pressing any one button, or
any two, or any three, or any four, or all five.
How many possible codes are there?

a)31
b)325
c)200
d)120
e)150
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 [#permalink] New post 14 Nov 2005, 19:41
Since order matters in this case:

5P1 + 5P2 + 5P3 + 5P4 + 5P5 = 325

I pick B.
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Re: PS-security code [#permalink] New post 15 Nov 2005, 01:43
It should be (5P1) + (5P2) + (5P3) + (5P4) + (5P5)
= 5 + 20 + 60 + 120 + 120 = 325
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 [#permalink] New post 15 Nov 2005, 09:19
yup you guys are right ..i came up with the same answer.

But OA was A 31 and OE was 2^5 -1 .

I was reading this question on a math website, I guess the answer on the site is wrong. its should be 325.

thanks
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code combination [#permalink] New post 15 Nov 2005, 23:28
cool_jonny009 wrote:
yup you guys are right ..i came up with the same answer.

But OA was A 31 and OE was 2^5 -1 .

I was reading this question on a math website, I guess the answer on the site is wrong. its should be 325.

thanks



cool_jonny009

I really hate all these Permutation/Compination formulas. Let think in simple logic way. As far as the combination of buttons pressed matters (I mean that combination 12 equals to combination 21) so we get only five possibilities for one button code pressed that are 1 or 2 or 3 or 4 or 5.

As for 5 button code we have - ONLY ONE possibility - all buttons pressed 1+2+3+4+5 (not 120 as permutation formula suggests).

Now lets think about 4 buttons combinations
1+2+3+4
1+2+3+5
1+2+4+5
1+3+4+5
2+3+4+5
Only five combinations.

I do not continue this boring counting, but for 2 buttons pressed we have 10 possible combinations, and for 3 buttons pressed we have 10 combinations also.

So, 5+10+10+5+1 gives us exactly 31. Sorry fot not using "rocket since approach" (permutation). I prefer old hacker's tools - paper and logic!

Have a nice day!

Mikki :)

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 [#permalink] New post 18 Nov 2005, 00:31
Mikki,

I see what you are saying..but what you have basically done is used combination instead of permutations, which others are using. You have made an assumption that the order of the keys entered would not make a difference, while the others have assumed otherwise.

From the question, it does not appear that the keys should be in order. But using our own knowledge we have assumed that security codes should be in order. Probably this is the trap.

Regards,

Alok Saboo
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 [#permalink] New post 18 Nov 2005, 07:22
Unfortunately the question is worded poorly. When I first read it, I took it to mean that you pushed the two buttons simultaneously therefore order doesn't matter which is what the question is saying (i.e. pushing 1 and 2 at the same time is the same as pushing 2 and 1 at the same time)

The only reason I thought of it this way is b/c I've used a lock like this before. I can see how this question would bring about confusion b/c most people think of pushing 1 then 2 instead of 1 and 2 at the same time.

The question should state more clearly that buttons are pushed at the same time if they want the answer to be 31.
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 [#permalink] New post 19 Nov 2005, 03:04
I got 325 and im sticking with it.

Its pretty annoying to solve such problems. Its common sense that when we operate a lock with numerical codes 21 is different from 12. How in heavens can we assume otherwise???

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 [#permalink] New post 19 Nov 2005, 04:46
Each number is different, therefore order matters and this is a permutations problem. Combinatorics problems deal with unordered arrangements (members in a committee or students in a study group).

Unless the problem is miscopied, the OA is 5P1 + 5P2 + 5P3 + 5P4 + 5P5 = 325
  [#permalink] 19 Nov 2005, 04:46
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