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Five letters are to be placed into five addressed envelopes.If the let

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Five letters are to be placed into five addressed envelopes.If the let  [#permalink]

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New post 14 May 2020, 21:39
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Five letters are to be placed into five addressed envelopes.If the letters are placed into the envelopes randomly, in how many ways can the letters be placed so that none of the letters is in its corresponding envelope?
A.44
B.30
C.120
D.80
E.85
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Re: Five letters are to be placed into five addressed envelopes.If the let  [#permalink]

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New post 14 May 2020, 23:19
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1
Number of derangement if there are total n objects \(= n!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+..........+(-1)^n \frac{1}{n!})\)

Number of ways in which none of the letters put in its corresponding envelope
\(= 5!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}) \)

\(= 120(\frac{1}{2} - \frac{1}{6} +\frac{1}{24} - \frac{1}{120})\)

\(= \frac{120}{120} (60-20+5-1)\)

=44


MBADream786 wrote:
Five letters are to be placed into five addressed envelopes.If the letters are placed into the envelopes randomly, in how many ways can the letters be placed so that none of the letters is in its corresponding envelope?
A.44
B.30
C.120
D.80
E.85
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Joined: 10 Jun 2018
Posts: 59
Re: Five letters are to be placed into five addressed envelopes.If the let  [#permalink]

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New post 14 May 2020, 23:41
To find the no. of cases in which none of the letters are placed in their corresponding envelopes, we have to subtract all the cases when at least one of the letter is correctly placed from the total no. of cases.
No letter is placed correctly = Total - (all letters are placed correctly) - (1 letter is placed correctly) - (2 letters are placed correctly) - (3 letters are placed correctly) - (4 letters are placed correctly)

Lets say there are letters La, Lb, Lc, Ld, Le and their corresponding envelopes Ea, Eb, Ec, Ed, Ee respectively.
Total no. of ways to assign each letter to its corresponding address = 5! = 5*4*3*2*1 = 120

1) All letters are placed correctly
All letters have 1 corresponding envelope. So each letter can be placed correctly in only 1 way.

2) 1 letter is placed correctly
Let’s first select the one letter out of 5 that must be put in its correct envelope. This can be done in 5C1 = 5 ways.
If 1 letter is placed correctly, we have to arrange remaining 4 letters in 4 envelopes such that they are not placed in their corresponding envelopes. Say La is placed in Ea. Now Lb cannot go into Eb. So Lb has remaining 3 envelopes as options. Similarly, Lc has remaining 3 options. Thus, all the remaining 4 letters each has 3options for envelopes such that they all are incorrectly placed.
Total ways in which 1 letter is placed correctly = 5*3 = 15

2) 2 letters are correctly placed
2 correctly placed letters can be selected in 5C2 = 10 ways
We have to place remaining 3 letters in 3 envelopes incorrectly. Since each letter cannot go in their respective addressed envelope, each letter has 2 options of envelopes (Since the third envelope would be their correct envelope).
Total no. of ways 2 letters are correctly placed = 10*2 = 20

3) 3 letters are correctly placed
3 correctly placed letters can be selected in 5C3 = 10 ways
We have to place remaining 2 letters in 2 envelopes incorrectly. Since each letter cannot go in their respective addressed envelope, each letter has just 1 option of envelope to go in(Since the other envelope would be their correct envelope).
Total no. of ways 3 letters are correctly placed = 10*1 = 10

4) 4 letters are correctly placed
This case is impossible since if 4 letters are put correctly in their envelopes, the last letter has to go in its designated envelope. So no. of ways = 0

Total no. of ways in which none of the letters are placed correctly = 120 - (1+15+20+10+0) = 74 ways

Where am I going wrong?
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Re: Five letters are to be placed into five addressed envelopes.If the let  [#permalink]

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New post 15 May 2020, 00:03
1
You calculated "1 letter is placed correctly" wrong

Suppose ABCDE is the right order

Number of ways to put only A in the right place

i)When B is at 3rd place - ACBED, ADBEC, AEBCD

ii)When B is at 4th place - ACEBD, ADEBC, AEDBC

iii) When B is at 5th place - ACEDB, ADECB, AEDCB

There are total 9 ways (Not 3)


2) 1 letter is placed correctly = 5C1*9 = 45


Total no. of ways in which none of the letters are placed correctly = 120 - (1+45+20+10+0) = 44 ways



abannore wrote:
To find the no. of cases in which none of the letters are placed in their corresponding envelopes, we have to subtract all the cases when at least one of the letter is correctly placed from the total no. of cases.
No letter is placed correctly = Total - (all letters are placed correctly) - (1 letter is placed correctly) - (2 letters are placed correctly) - (3 letters are placed correctly) - (4 letters are placed correctly)

Lets say there are letters La, Lb, Lc, Ld, Le and their corresponding envelopes Ea, Eb, Ec, Ed, Ee respectively.
Total no. of ways to assign each letter to its corresponding address = 5! = 5*4*3*2*1 = 120

1) All letters are placed correctly
All letters have 1 corresponding envelope. So each letter can be placed correctly in only 1 way.

2) 1 letter is placed correctly
Let’s first select the one letter out of 5 that must be put in its correct envelope. This can be done in 5C1 = 5 ways.
If 1 letter is placed correctly, we have to arrange remaining 4 letters in 4 envelopes such that they are not placed in their corresponding envelopes. Say La is placed in Ea. Now Lb cannot go into Eb. So Lb has remaining 3 envelopes as options. Similarly, Lc has remaining 3 options. Thus, all the remaining 4 letters each has 3options for envelopes such that they all are incorrectly placed.
Total ways in which 1 letter is placed correctly = 5*3 = 15

2) 2 letters are correctly placed
2 correctly placed letters can be selected in 5C2 = 10 ways
We have to place remaining 3 letters in 3 envelopes incorrectly. Since each letter cannot go in their respective addressed envelope, each letter has 2 options of envelopes (Since the third envelope would be their correct envelope).
Total no. of ways 2 letters are correctly placed = 10*2 = 20

3) 3 letters are correctly placed
3 correctly placed letters can be selected in 5C3 = 10 ways
We have to place remaining 2 letters in 2 envelopes incorrectly. Since each letter cannot go in their respective addressed envelope, each letter has just 1 option of envelope to go in(Since the other envelope would be their correct envelope).
Total no. of ways 3 letters are correctly placed = 10*1 = 10

4) 4 letters are correctly placed
This case is impossible since if 4 letters are put correctly in their envelopes, the last letter has to go in its designated envelope. So no. of ways = 0

Total no. of ways in which none of the letters are placed correctly = 120 - (1+15+20+10+0) = 74 ways

Where am I going wrong?
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Re: Five letters are to be placed into five addressed envelopes.If the let  [#permalink]

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New post 15 May 2020, 20:48
nick1816 wrote:
You calculated "1 letter is placed correctly" wrong

Suppose ABCDE is the right order

Number of ways to put only A in the right place

i)When B is at 3rd place - ACBED, ADBEC, AEBCD

ii)When B is at 4th place - ACEBD, ADEBC, AEDBC

iii) When B is at 5th place - ACEDB, ADECB, AEDCB

There are total 9 ways (Not 3)


2) 1 letter is placed correctly = 5C1*9 = 45


Total no. of ways in which none of the letters are placed correctly = 120 - (1+45+20+10+0) = 44 ways



abannore wrote:
To find the no. of cases in which none of the letters are placed in their corresponding envelopes, we have to subtract all the cases when at least one of the letter is correctly placed from the total no. of cases.
No letter is placed correctly = Total - (all letters are placed correctly) - (1 letter is placed correctly) - (2 letters are placed correctly) - (3 letters are placed correctly) - (4 letters are placed correctly)

Lets say there are letters La, Lb, Lc, Ld, Le and their corresponding envelopes Ea, Eb, Ec, Ed, Ee respectively.
Total no. of ways to assign each letter to its corresponding address = 5! = 5*4*3*2*1 = 120

1) All letters are placed correctly
All letters have 1 corresponding envelope. So each letter can be placed correctly in only 1 way.

2) 1 letter is placed correctly
Let’s first select the one letter out of 5 that must be put in its correct envelope. This can be done in 5C1 = 5 ways.
If 1 letter is placed correctly, we have to arrange remaining 4 letters in 4 envelopes such that they are not placed in their corresponding envelopes. Say La is placed in Ea. Now Lb cannot go into Eb. So Lb has remaining 3 envelopes as options. Similarly, Lc has remaining 3 options. Thus, all the remaining 4 letters each has 3options for envelopes such that they all are incorrectly placed.
Total ways in which 1 letter is placed correctly = 5*3 = 15


2) 2 letters are correctly placed
2 correctly placed letters can be selected in 5C2 = 10 ways
We have to place remaining 3 letters in 3 envelopes incorrectly. Since each letter cannot go in their respective addressed envelope, each letter has 2 options of envelopes (Since the third envelope would be their correct envelope).
Total no. of ways 2 letters are correctly placed = 10*2 = 20

3) 3 letters are correctly placed
3 correctly placed letters can be selected in 5C3 = 10 ways
We have to place remaining 2 letters in 2 envelopes incorrectly. Since each letter cannot go in their respective addressed envelope, each letter has just 1 option of envelope to go in(Since the other envelope would be their correct envelope).
Total no. of ways 3 letters are correctly placed = 10*1 = 10

4) 4 letters are correctly placed
This case is impossible since if 4 letters are put correctly in their envelopes, the last letter has to go in its designated envelope. So no. of ways = 0

Total no. of ways in which none of the letters are placed correctly = 120 - (1+15+20+10+0) = 74 ways

Where am I going wrong?


Hi Nick,
Could you please explain why the logic here is not working? The same procedure works for the other scenarios, I'm unable to arrive at 5*9 using the logic that works for the other scenarios.
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Re: Five letters are to be placed into five addressed envelopes.If the let  [#permalink]

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New post 16 May 2020, 00:32
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Logic is not correct at all, that's why it's working for only few cases.


If we put only first 2 letter in correct envelopes, that doesn't mean that we have 2 options for each of the remaining spots (If it were, then our answer would be 2*2*2=8 not 2) , because filling each spot is dependent on other in this question.(We need inclusion-exclusion principle to calculate)

Suppose ABCDE is the correct order. Let's see the cases in which only AB is at correct place.


Case 1- when C is at 4th place - ABECD (1 arrangement)
Case 2- when C is at 5th place- ABDEC (1 arrangement)

Total possible arrangements 1+1=2(right way to solve it)

Since there is only 1 arrangement possible in each case, that false logic works(2*1).If the arrangements are more than 1 in each case, then this logic fails(as in the following case.)

Suppose ABCDE is the right order

Number of ways to put only A in the right place

i)When B is at 3rd place - ACBED, ADBEC, AEBCD

ii)When B is at 4th place - ACEBD, ADEBC, AEDBC

iii) When B is at 5th place - ACEDB, ADECB, AEDCB

We have 3 arrangements possible in each case. (total possible arrangements is 3+3+3, not 3*1)


If you still have any doubt, you can ask.



ShreyasJavahar wrote:
Hi Nick,
Could you please explain why the logic here is not working? The same procedure works for the other scenarios, I'm unable to arrive at 5*9 using the logic that works for the other scenarios.
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Re: Five letters are to be placed into five addressed envelopes.If the let   [#permalink] 16 May 2020, 00:32

Five letters are to be placed into five addressed envelopes.If the let

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