A secretary types 4 letters and then addresses the 4 corresponding

If 3 letter are correct, the fourth one has no other option but to go into the right one. So, the no.of cases where only 3 letters are correct is 0

One can simply solve these kind of questions by the concept of Derangements.

Derangement is given by the formula:

\(D_n\) = \({1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - ..... (-1)^n \frac{1}{n!}}\)

So in this case, we have to derange the 4 cards as no card is going in its correct envelope.

So \(D_4\) should be calculated.

\(D_4\) = \({1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!}}\)

\(D_4\) = \(\frac{1}{2} - \frac{1}{6} + \frac{1}{24}\)

\(D_4\) = \(\frac{9}{24}\)

So out of 24 total cases (4!), there will be 9 cases where no card goes into the correct envelope.

OPTION: B

For more info on the concept of derangements please refer to the Experts' Global concept video:

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

ASIDE: Here's a similar question to practice with: https://gmatclub.com/forum/tanya-prepar ... 85167.html

Cheers,
Brent

GMATPrepNow is this correct reasoning?

The mistake I made was in my break down of the problem: I thought that "any letter can go into any box besides it's own, so 3 out of 4 possibilities * 4 boxes (aA aB, aC, aD, bA, bB, bC, bD, etc... so 3*4 = 12)" This assumes replacement, which we are not doing in this problem.

If I understand right, it's more complicated because not only can the first letter not go into its own box, but the second can't go into its own, and so on... so we have to fulfill the constraint for every step, which results in less viable possibilities (should be 4! total... i.e. starting with each letter makes 1*3*2*1 = 6 possibilities * 4 letters, not 4^4 as I thought)

This was a lot more clear when I made a table. Everything that starts with A is out, and for the other 3 letters there are 3/6 possibilities. Some of them work up to a certain point. For example starting with B, you can fulfill the requirement for the first 3 slots with bcad, but not the fourth slot by having d in the fourth slot (corresponding to box D makes it nonviable) So it's actually 9/24.

This concept has been explained really well by VeritasKarishma.

From Karishma's blog: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/blog/2011/ ... envelopes/.

La can be put in either Eb or Ec or Ed (i.e. 3 ways). Say, La is put in Ec. Now we have 3 letters leftover: Lb, Lc and Ld and 3 envelopes leftover: Ea, Eb and Ed. Lc, the letter corresponding to Ec, can be put in any one of these three envelopes. Hence Lc can be put in 3 ways too. Say, Lc is put in Ed. Now, we have two letters, Lb and Ld leftover and two envelopes, Ea and Eb leftover. Lb cannot go into Eb so Lb must go into Ea and Ld must go into Eb i.e. there is only one way of putting in the other two letters. So, number of ways of putting in all the letters incorrectly = 3*3*1 = 9 ways.

Assume the envelope positions are "locked" so that you are just arranging four letters among the four positions.
Total number of ways to arrange four letters in four positions is 4! = 24

Now to subtract the number of ways in which one or more letters are in their right envelopes.

4 letters correct = 4C4 = 1
3 letters correct = 4C3 = 4
2 letters correct = 4C2 = 6
1 letter correct = 4C1 = 4

So 15 different ways in which at least 1 letter is put in the right envelope, so 24 - 15 = 9 ways in which ZERO letters are in the right envelope

If 1 letter is correct, then remaining 3 letters may be put in 2 incorrect ways:
Correct places: ABCD
Number of options in which only A is correctly placed - 2.

ADBC, ACDB.

So there are 2*4=8 options if 1 letter is correct.

Does it make sense?

Posted from my mobile device

I am confused about this approach.
How can we have a case in which 3 letters are correct?
No of letters and no of envelopes are equal. If 3 are correct, 4th will definitely be correct.

Same question. Shouldn't the number of combinations of 3 letters correct and 4 letters correct be same ? You can't go wrong with 1 and get the other 3 letters correct.

Bunuel chetan2u

This is answered here: https://gmatclub.com/forum/a-secretary- ... l#p2420957
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Question)
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?


Solution:
Let the letters be A, B, C and D; and the envelopes be a, b, c, and d
Where A goes to a, B to b, C to c, and D to d (in the ideal case)

However, we want all letters wrongly assigned.
Let us start from A:
A can be assigned to: b or c or d
Let A be assigned to b:
Remaining letters: B, C, D and remaining envelopes: a, c, d
Case 1: B goes to a => C must go to d and D must go to c ..... (1 way)
Case 2: B goes to c => C must go to d and D must go to a ..... (1 way)
Case 3: B goes to d => C must go to a and D must go to c ..... (1 way)
Thus, when A is assigned to b, we have 3 cases
We would also have 3 cases for each of the cases when A is assigned to c, or when A is assigned to d
Thus, total ways = 3 + 3 + 3 = 9


Alternate Method:

Let the letters be A, B, C and D; and the envelopes be a, b, c, and d
Where A goes to a, B to b, C to c, and D to d (in the ideal case)
If we randomly assign letters to envelopes, we can do that in 4! = 24 possible ways.

There are 4 possible cases:
1. All letters assigned correctly - 1 way
2. One letter assigned correctly -
Let the letter A be assigned correctly to a.
We are left with B, C, D and envelopes b, c, d
We need to assign these wrongly: there are 2 ways: B - c, C - d, D - b OR B - d, C - b, D - c
Thus, if B is assigned correctly to b, then too we will have 2 cases; similarly for C and D
Thus, there are a total of 8 ways when one letter is assigned correctly
3. Two letters assigned correctly -
We can select the 2 letters we want to assign correctly in 4C2 = 6 ways
The remaining 2 letters must be assigned to the other's envelope.
For example, if C and D need to be wrongly assigned, it must be that C goes to d and D to c - there is only 1 way
Thus, there are a total of 6 ways when two letters are assigned correctly
4. All letters are wrongly assigned (Note: you cannot have 3 letters assigned correctly and 1 incorrectly) -
Required number = 4! - (1 + 8 + 6) = 9 ways

Answer B


Note: This can also be done by the method of derangements. However, I feel that, for the GMAT, knowing this concept is not necessary
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Hi, I have unmarked the solution as best reply as it is wrong on two counts..
1) Firstly, as observed by you, there can be no way that 3 are correct and one is incorrect.
2) Secondly, 4! means Permutation, that is order matters, where as 4C1, 4C2 etc means combination, that is order does NOT matter. You cannot subtract one from the other.

For the solution part if I were to follow the same method.
1) All 4 correct - 4C4*1=1
2) Only 3 correct -- Not possible
3) Only 2 correct -- 4C2*1=6... Now when you have placed two correctly, the other two can be placed in only 1 way.. AB are placed correctly, so C can have only D and D can have only C.
4) Only 1 correct -- 4C1*2=8.. When one is correct, the other 3 can be placed incorrectly in 2 ways.. ..Say A is placed correctly, B can have C or D, and C & D can have only 1 envelope.
B has C, C will have D and D will have B OR B has D, C will have B and D will have C.
Total = 1+0+6+8=15..
Our answer 4!-15=24-15=9.

ANOTHER way..
4 incorrect..
Say A has B, so B can have any of A, C or D..
---B has A, C and D can interchange--1 way
---B has C, C will have D and D, A -- 1 way
---B has D, C will have A and D, C -- 1 way
Thus total 3 ways.
Similarly A can have C, then 3 ways and A can have D, again 3 ways..
Total 3+3+3=9 ways.

Derangement formula
A derangement is a permutation in which none of the objects appear in their "natural" place.
Formula is \(n!(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}....+\frac{(-1)^n}{n!})\)

Here n is 4, so answer = \(4!(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!})=4*3-4+1=12-4+1=9ways\)

B
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How can 3 letters be correctly placed but the 4th one placed incorrectly? The value for 3 letters correct should be 0, no?

We want that no letter should go to the correct envelope.

Simple apply de-arrangement:--

4!(1/2!-1/3!-1/4!)
=24((12-4+1)/24)
=9


IMO B.

For letters related such questions,de-arrangement is very time saving and you can solve hard questions in a minute.

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