A secretary types 4 letters and then addresses the 4 : PS Archive
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 19 Jan 2017, 09:44

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# A secretary types 4 letters and then addresses the 4

Author Message
Manager
Joined: 22 Oct 2007
Posts: 120
Followers: 1

Kudos [?]: 59 [0], given: 0

### Show Tags

27 Apr 2008, 04:43
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
Intern
Joined: 26 Apr 2008
Posts: 49
Followers: 1

Kudos [?]: 2 [0], given: 0

Re: PS - Permutations - 4 letters and 4 envelopes [#permalink]

### Show Tags

27 Apr 2008, 08:00
I got 15. Lemme know it its correct and I'll explain if it is.
Senior Manager
Joined: 19 Apr 2008
Posts: 320
Followers: 3

Kudos [?]: 78 [0], given: 0

Re: PS - Permutations - 4 letters and 4 envelopes [#permalink]

### Show Tags

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
Intern
Joined: 26 Apr 2008
Posts: 49
Followers: 1

Kudos [?]: 2 [0], given: 0

Re: PS - Permutations - 4 letters and 4 envelopes [#permalink]

### Show Tags

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.
Senior Manager
Joined: 19 Apr 2008
Posts: 320
Followers: 3

Kudos [?]: 78 [0], given: 0

Re: PS - Permutations - 4 letters and 4 envelopes [#permalink]

### Show Tags

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
Senior Manager
Joined: 10 Mar 2008
Posts: 371
Followers: 5

Kudos [?]: 259 [0], given: 0

Re: PS - Permutations - 4 letters and 4 envelopes [#permalink]

### Show Tags

27 Apr 2008, 19:34
Could one of you please tell me the source of this question?
Intern
Joined: 26 Apr 2008
Posts: 49
Followers: 1

Kudos [?]: 2 [0], given: 0

Re: PS - Permutations - 4 letters and 4 envelopes [#permalink]

### Show Tags

27 Apr 2008, 22:21
Looks right. Seems that I am making some mistake..
Manager
Joined: 22 Oct 2007
Posts: 120
Followers: 1

Kudos [?]: 59 [0], given: 0

Re: PS - Permutations - 4 letters and 4 envelopes [#permalink]

### Show Tags

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.
Senior Manager
Joined: 19 Apr 2008
Posts: 320
Followers: 3

Kudos [?]: 78 [0], given: 0

Re: PS - Permutations - 4 letters and 4 envelopes [#permalink]

### Show Tags

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.
Intern
Joined: 26 Apr 2008
Posts: 49
Followers: 1

Kudos [?]: 2 [0], given: 0

Re: PS - Permutations - 4 letters and 4 envelopes [#permalink]

### Show Tags

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
Joined: 19 Apr 2008
Posts: 320
Followers: 3

Kudos [?]: 78 [0], given: 0

Re: PS - Permutations - 4 letters and 4 envelopes [#permalink]

### Show Tags

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
Display posts from previous: Sort by