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# A secretary types 4 letters and then addresses the 4

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A secretary types 4 letters and then addresses the 4 [#permalink]  27 Apr 2008, 04:43
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A secretary types 4 letters and then addresses the 4 corresponding envelopes. In how many ways can the secretary place the letters in the envelopes so that NO letter is placed in its correct envelope?

a. 8
b. 9
c. 10
d. 12
e. 15
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Re: PS - Permutations - 4 letters and 4 envelopes [#permalink]  27 Apr 2008, 08:00
I got 15. Lemme know it its correct and I'll explain if it is.
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Re: PS - Permutations - 4 letters and 4 envelopes [#permalink]  27 Apr 2008, 11:33
A secretary types 4 letters and then addresses the 4 corresponding envelopes. In how many ways can the secretary place the letters in the envelopes so that NO letter is placed in its correct envelope?

a. 8
b. 9
c. 10
d. 12
e. 15

my answer is 9 . (3+3+3) is that right?

though I can answer the probability of secretary getting fired, if this happened ..it would be straight 1 wish GMAT would ask questions like this ..haha
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Re: PS - Permutations - 4 letters and 4 envelopes [#permalink]  27 Apr 2008, 11:48
Cant be 3+3+3. Try doing it.

A B C D
a b c d

write the lowercase alphabets below the upper case so that no pair has the same alphabets.
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Re: PS - Permutations - 4 letters and 4 envelopes [#permalink]  27 Apr 2008, 12:33
itsme291 wrote:
Cant be 3+3+3. Try doing it.

A B C D
a b c d

write the lowercase alphabets below the upper case so that no pair has the same alphabets.

even with that method my answer is 9 , am I missin something

A B C D
b c d a
b a d c
b d a c
c a d b
c d b a
c d a b
d a b c
d c b a
d c a b
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Re: PS - Permutations - 4 letters and 4 envelopes [#permalink]  27 Apr 2008, 19:34
Could one of you please tell me the source of this question?
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Re: PS - Permutations - 4 letters and 4 envelopes [#permalink]  27 Apr 2008, 22:21
Looks right. Seems that I am making some mistake..
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Re: PS - Permutations - 4 letters and 4 envelopes [#permalink]  27 Apr 2008, 23:37
rpmodi wrote:
itsme291 wrote:
Cant be 3+3+3. Try doing it.

A B C D
a b c d

write the lowercase alphabets below the upper case so that no pair has the same alphabets.

even with that method my answer is 9 , am I missin something

A B C D
b c d a
b a d c
b d a c
c a d b
c d b a
c d a b
d a b c
d c b a
d c a b

Yes Mr. Modi..u r bang on target. Could u please explain the approach of reaching the ans without listing down all options? This listing down options business is too time consuming.
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Re: PS - Permutations - 4 letters and 4 envelopes [#permalink]  28 Apr 2008, 09:31
rpmodi wrote:
itsme291 wrote:
Cant be 3+3+3. Try doing it.

A B C D
a b c d

write the lowercase alphabets below the upper case so that no pair has the same alphabets.

even with that method my answer is 9 , am I missin something

A B C D
b c d a
b a d c
b d a c
c a d b
c d b a
c d a b
d a b c
d c b a
d c a b

Yes Mr. Modi..u r bang on target. Could u please explain the approach of reaching the ans without listing down all options? This listing down options business is too time consuming.

diffrent approaches here

1) First as I have listed all the combination , just list the numbers for one letter being the wrong envelop and then pattern repeats so you can simply add and that's how I got 3+3+3

For e.g in the above example , I just have to jot down different combinations for letter b in envelop A , since I got 3 combinations , pattern will repeat for c and d . (obviously you have to rule out all the combinations of letter a in envelop A )

A B C D
b c d a
b a d c
b d a c

Itsme's answer would have been correct if it were asked what are the different ways that atleast one letter is in correct envelop .

Last edited by rpmodi on 28 Apr 2008, 09:44, edited 1 time in total.
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Re: PS - Permutations - 4 letters and 4 envelopes [#permalink]  28 Apr 2008, 09:38
Yes. rpmodi is right.
Actually my approach was to subtract the cases when atleast one envelope-letter combo is right.
4!-15 = 9

Forgot to do the last part in hurry
Senior Manager
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Re: PS - Permutations - 4 letters and 4 envelopes [#permalink]  28 Apr 2008, 10:03
itsme291 wrote:
Yes. rpmodi is right.
Actually my approach was to subtract the cases when atleast one envelope-letter combo is right.
4!-15 = 9

Forgot to do the last part in hurry

I have generalized the formula for this kind of problem

(n!-(n-1)!)/2 ; caution: verify before applying or use it at your own risk
Re: PS - Permutations - 4 letters and 4 envelopes   [#permalink] 28 Apr 2008, 10:03
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