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A secretary types 4 letters and then addresses the 4
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Updated on: 23 Jul 2019, 04:47
Updated on: 23 Jul 2019, 04:47
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Question Stats:
38% (01:45) correct
61% (01:54) wrong based on 109 sessions
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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
A) 8
B) 9
C) 10
D) 12
E) 15
Re: A secretary types 4 letters and then addresses the 4
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10 Sep 2019, 20:07
10 Sep 2019, 20:07
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By the wording, if even 1 letter is in the correct envelope, it counts as a fail.
In all, there are 4!=(4*3*2*1) or 24 possible ways to put the four letters into 4 different envelopes. Now we need to subtract the failures.
Failure will occur if exactly 1 letter is in the correct envelope. Since there are 4 envelops, there are 4 different ways to accomplish this. So, the value is 4 chose 1. Likewise, there are 4 choose 2 ways to insert 2 letters into the correct envelopes, 4 choose 3 ways to insert 3 letters into the 4 envelopes, and 4 chose 4 ways to insert all 4 letters into the correct envelopes.
Let n=4 be the number of envelopes in which at least 1 of n letters are to be incorrectly placed and let m be the total number of ways of doing so. Then, m can be expressed as below.
m=24−[(4C1)+(4C2)+(4C3)+(4C4)]
m=24−[4+6+4+1]=24−15=9
So, there are 9 ways in which the 4 letters can be inserted into the 4 different envelops in which at least 1 is incorrect.
In all, there are 4!=(4*3*2*1) or 24 possible ways to put the four letters into 4 different envelopes. Now we need to subtract the failures.
Failure will occur if exactly 1 letter is in the correct envelope. Since there are 4 envelops, there are 4 different ways to accomplish this. So, the value is 4 chose 1. Likewise, there are 4 choose 2 ways to insert 2 letters into the correct envelopes, 4 choose 3 ways to insert 3 letters into the 4 envelopes, and 4 chose 4 ways to insert all 4 letters into the correct envelopes.
Let n=4 be the number of envelopes in which at least 1 of n letters are to be incorrectly placed and let m be the total number of ways of doing so. Then, m can be expressed as below.
m=24−[(4C1)+(4C2)+(4C3)+(4C4)]
m=24−[4+6+4+1]=24−15=9
So, there are 9 ways in which the 4 letters can be inserted into the 4 different envelops in which at least 1 is incorrect.
General Discussion
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Re: A secretary types 4 letters and then addresses the 4
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23 Jul 2019, 04:43
23 Jul 2019, 04:43
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huda wrote:
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) \(15\)
A) \(8\)
B) \(9\)
C) \(10\)
D) \(12\)
E) \(15\)
Show: :: OA
Confused with answer 9 and 12. 
When we scan the answer choices (ALWAYS scan the answer choices before choose your plan of attack), we see that all of the answer choices are relatively small.
So, a perfectly valid approach is to list and count the possible outcomes
Let a, b, c and d represent the letters, and let A, B, C and D represent the corresponding addresses.
So, let's list the letters in terms of the order in which they are delivered to addresses A, B, C, and D.
So, for example, the outcome abcd would represent all letters going to their intended addresses.
Likewise, cabd represent letter d going to its intended address, but the other letters not going to their intended addresses.
Now let's list all possible outcomes where ZERO letters go to their intended addresses:
- badc
- bcda
- bdac
- cadb
- cdab
- cdba
- dabc
- dcab
- dcba
DONE!
There are 9 such outcomes.
Answer: B
Cheers,
Brent
