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A secretary types 4 letters and then addresses the 4 [#permalink]
23 Jul 2019, 04:42
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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
Last edited by huda on 23 Jul 2019, 04:47, edited 4 times in total.




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Re: A secretary types 4 letters and then addresses the 4 [#permalink]
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) \(15\) 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
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Re: A secretary types 4 letters and then addresses the 4 [#permalink]
23 Jul 2019, 06:30
This is problemsolving NOT a numeric entry question. Moreover, use the timer for the Official Answer. regards
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Re: A secretary types 4 letters and then addresses the 4 [#permalink]
24 Jul 2019, 07:09
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total = 4! = 24 one letter placed correctly= 4C1*2 = 8 two letter place correctly = 4C2*1 = 6 three/four letter placed correctly = 1 no letter placed correctly = 24  (8+6+1) = 9



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Re: A secretary types 4 letters and then addresses the 4 [#permalink]
22 Aug 2019, 17:09
shubhpan wrote: total = 4! = 24 one letter placed correctly= 4C1*2 = 8 two letter place correctly = 4C2*1 = 6 three/four letter placed correctly = 1 no letter placed correctly = 24  (8+6+1) = 9 Would you mind helping me understand where the "*2" and the "*1" come from?



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Re: A secretary types 4 letters and then addresses the 4 [#permalink]
23 Aug 2019, 00:29
Unclear also to me.
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Re: A secretary types 4 letters and then addresses the 4 [#permalink]
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.



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Re: A secretary types 4 letters and then addresses the 4 [#permalink]
11 Sep 2019, 05:30
I had a different approach to solve this problem. Let us consider the different combinations in which the letters be placed in incorrect envelopes. We have 4 letters.. For the 1st letter, there are 3 incorrect envelopes to choose from Similarly for the 2nd and 3rd letter there are again 3 incorrect envelopes to chose from By this time, the 4th letter will automatically have the wrong envelope choice left.
Therefore, the number of combinations = 3*3*3*1 = 9



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Re: A secretary types 4 letters and then addresses the 4 [#permalink]
11 Sep 2019, 06:04
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agodbole wrote: I had a different approach to solve this problem. Let us consider the different combinations in which the letters be placed in incorrect envelopes. We have 4 letters.. For the 1st letter, there are 3 incorrect envelopes to choose from Similarly for the 2nd and 3rd letter there are again 3 incorrect envelopes to chose from By this time, the 4th letter will automatically have the wrong envelope choice left.
Therefore, the number of combinations = 3*3*3*1 = 9 This is a good idea, but there are a few problems with that approach. Let a, b, c and d represent the letters, and let A, B, C and D represent the corresponding addresses. Step 1: Choose an envelope for letter a. We can do this in 3 ways. Step 2: Choose an envelope for letter b. If, in step 1, we placed letter a in address B, then we can place b in 3 different addresses (A, C or D) If, in step 1, we placed letter a in address C, then we can place b in 2 different addresses (A or D) So, which is it? 3 or 2? The problem exists for steps 3 and 4 as well. Also, 3*3*3*1 = 27 (not 9) Cheers, Brent
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Re: A secretary types 4 letters and then addresses the 4 [#permalink]
11 Sep 2019, 06:15
GreenlightTestPrep wrote: agodbole wrote: I had a different approach to solve this problem. Let us consider the different combinations in which the letters be placed in incorrect envelopes. We have 4 letters.. For the 1st letter, there are 3 incorrect envelopes to choose from Similarly for the 2nd and 3rd letter there are again 3 incorrect envelopes to chose from By this time, the 4th letter will automatically have the wrong envelope choice left.
Therefore, the number of combinations = 3*3*3*1 = 9 This is a good idea, but there are a few problems with that approach. Let a, b, c and d represent the letters, and let A, B, C and D represent the corresponding addresses. Step 1: Choose an envelope for letter a. We can do this in 3 ways. Step 2: Choose an envelope for letter b. If, in step 1, we placed letter a in address B, then we can place b in 3 different addresses (A, C or D) If, in step 1, we placed letter a in address C, then we can place b in 2 different addresses (A or D) So, which is it? 3 or 2? The problem exists for steps 3 and 4 as well. Also, 3*3*3*1 = 27 (not 9) Cheers, Brent OMG I can't math! Sorry.. been a bad morning thusfar..



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Re: A secretary types 4 letters and then addresses the 4 [#permalink]
11 Sep 2019, 08:30
agodbole wrote: OMG I can't math! Sorry.. been a bad morning thus far.. Go easy on yourself. It's a very tricky question (38% correct) Cheers, Brent
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Re: A secretary types 4 letters and then addresses the 4
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