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A secretary types 4 letters and then addresses the 4 corresponding

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Senior Manager
Status: Gathering chakra
Joined: 05 Feb 2018
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A secretary types 4 letters and then addresses the 4 corresponding  [#permalink]

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11 Mar 2019, 13:57
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Difficulty:

95% (hard)

Question Stats:

33% (01:42) correct 67% (01:49) wrong based on 118 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

Official answer in spoiler below, anyone have a better explanation/approach?

The question stem asks how many ways, indicating that this is a combination/permutation question. First, recognize that this is a permutation question because the order of the letters to the envelopes matters. Unfortunately, the restriction that NO letter is placed in its correct envelope is so restrictive that it will be easier to list out the possibilities than to calculate them. Start with a single scenario and then extrapolate. For discussion, designate the letters L1, L2, L3 and L4, and designate the envelopes E1, E2, E3, and E4.

First, since L1 cannot be assigned to E1, assign L1 to E2. Then list out the possibilities for the remaining three letters such that NO letter is placed in its correct envelope. The only three possibilities are (L2 to E1, L3 to E4, L4 to E3), (L2 to E3, L3 to E4, L4 to E1), and (L2 to E4, L3 to E1, L4 to E3). There are 3 possible ways to assign the remaining letters when L1 assigned to E2. So, there must be 3 possible ways to assign the remaining letters with L1 assigned to E3, and there must be 3 possible ways to assign the remaining letters with L1 assigned to E4. Therefore, there are a total of 3 + 3 + 3 = 9 ways the secretary can place the letters in the envelopes so that NO letter is placed in its correct envelope.

The correct answer is choice B.
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Re: A secretary types 4 letters and then addresses the 4 corresponding  [#permalink]

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17 Mar 2019, 20:17
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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
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A secretary types 4 letters and then addresses the 4 corresponding  [#permalink]

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11 Mar 2019, 17:52
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energetics 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

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:
- bcda
- bdac
- cdab
- cdba
- dabc
- dcab
- dcba

DONE!
There are 9 such outcomes.

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

Cheers,
Brent
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A secretary types 4 letters and then addresses the 4 corresponding  [#permalink]

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12 Mar 2019, 06:53
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.
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A secretary types 4 letters and then addresses the 4 corresponding  [#permalink]

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15 Mar 2019, 06:40
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energetics 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

Official answer in spoiler below, anyone have a better explanation/approach?

The question stem asks how many ways, indicating that this is a combination/permutation question. First, recognize that this is a permutation question because the order of the letters to the envelopes matters. Unfortunately, the restriction that NO letter is placed in its correct envelope is so restrictive that it will be easier to list out the possibilities than to calculate them. Start with a single scenario and then extrapolate. For discussion, designate the letters L1, L2, L3 and L4, and designate the envelopes E1, E2, E3, and E4.

First, since L1 cannot be assigned to E1, assign L1 to E2. Then list out the possibilities for the remaining three letters such that NO letter is placed in its correct envelope. The only three possibilities are (L2 to E1, L3 to E4, L4 to E3), (L2 to E3, L3 to E4, L4 to E1), and (L2 to E4, L3 to E1, L4 to E3). There are 3 possible ways to assign the remaining letters when L1 assigned to E2. So, there must be 3 possible ways to assign the remaining letters with L1 assigned to E3, and there must be 3 possible ways to assign the remaining letters with L1 assigned to E4. Therefore, there are a total of 3 + 3 + 3 = 9 ways the secretary can place the letters in the envelopes so that NO letter is placed in its correct envelope.

The correct answer is choice B.

This concept has been explained really well by VeritasKarishma.

From Karishma's blog: https://www.veritasprep.com//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.
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Re: A secretary types 4 letters and then addresses the 4 corresponding  [#permalink]

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17 Jul 2019, 21:29
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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

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Re: A secretary types 4 letters and then addresses the 4 corresponding  [#permalink]

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09 Dec 2019, 15:53
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siyeezy wrote:
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 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
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Re: A secretary types 4 letters and then addresses the 4 corresponding  [#permalink]

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10 Dec 2019, 06:27
AnirudhChalla wrote:
siyeezy wrote:
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 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

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.

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

Does it make sense?

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A secretary types 4 letters and then addresses the 4 corresponding  [#permalink]

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14 Dec 2019, 22:00
siyeezy wrote:
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

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.
A secretary types 4 letters and then addresses the 4 corresponding   [#permalink] 14 Dec 2019, 22:00
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