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# A bank teller wants to open a safe at the bank. However, she

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Director
Joined: 17 Oct 2005
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A bank teller wants to open a safe at the bank. However, she [#permalink]  11 Feb 2006, 12:44
A bank teller wants to open a safe at the bank. However, she forget the pin number. She does not want to ask anyone because she doesnt want anyone to think that she is incompetent. She remembers that the pin number consists of a 5 digit number and was formed from the digits 1,2,3,4, and 5. She also remembers that no digit appears more than once and that 1 and 5 cannot be adjacent. How many such numbers are possible?
VP
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conocieur wrote:
5! - 4! = 96

conocieur did you consider 15 and 51 as two possibilities? Check your solution again!
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Manager
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Is it - 72

5! - 4! *2 = 72
Director
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72 is the oa.

Can you explain more indepthly? thanks!
Manager
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Re: codes [#permalink]  12 Feb 2006, 07:02
joemama142000 wrote:
A bank teller wants to open a safe at the bank. However, she forget the pin number. She does not want to ask anyone because she doesnt want anyone to think that she is incompetent. She remembers that the pin number consists of a 5 digit number and was formed from the digits 1,2,3,4, and 5. She also remembers that no digit appears more than once and that 1 and 5 cannot be adjacent. How many such numbers are possible?

The way I looked at it is this:

the no. of arrangements of numbers 1- 5 in which 1 and 5 are not adjacent= Total no. of arrangements of the 5 nos. - the no. of arrangements in which 1 & 5 are adjacent.
Now, Total no. of arrangements of the 5 nos.= 5!
and the no. of arrangements in which 1 & 5 are adjacent. = 4!*2 (because- consider 1 & 5 as one no., always together, then we would have 4 nos. to arrange, so 4!. But also in considering 1 & 5 as one number we still have two possibilites of arranging 1 & 5 ie. 15 or 51, hence the multiplication by 2).
Hope it helps.
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giddi77 wrote:
conocieur wrote:
5! - 4! = 96

conocieur did you consider 15 and 51 as two possibilities? Check your solution again!

you're right

5! - 2*4! = 72
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# of permutations with no constraints = 5! = 120
# of permutations with 1 and 5 beside each other = 4!*2 = 24*2 = 48
# of permutations where 1 and 5 are not beside each other = 120-48 = 72
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