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A secretary types 4 letters and then addresses the 4

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

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New post 25 Aug 2008, 19:07
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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

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

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New post 25 Aug 2008, 19:42
the ans is 12 ie D

the correct arrangement of letter is 1 1 1 1
wrongly placed is it can be done by3 3 3 3 ways so 4*3 = 12 ways

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

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New post 26 Aug 2008, 00:12
vtselvan wrote:
the ans is 12 ie D

the correct arrangement of letter is 1 1 1 1
wrongly placed is it can be done by3 3 3 3 ways so 4*3 = 12 ways


could you expand on that?
what do u mean by wrongly placed is it can be done by3 3 3 3 ?

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

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New post 26 Aug 2008, 00:54
ssandeepan 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


Total number of ways = 3 + 3 + 3 + 3 = 12

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

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New post 26 Aug 2008, 03:51
I was checking if anybody has an easier explanation . Otherwise it goes like this :

answer = total number of ways of arranging letters - (one letter in right envelope + 2 letters in right envelope+ 3 letters in right envelope + 4 letters in right envelope)

Now,
Let the no. of ways in which exactly 1 letter goes into the right envelope & rest of the 3 letters go into 3 wrong envelopes be x1
Let the no. of ways in which exactly 2 letters go into the right envelopes & rest of the 2 letters go into 2 wrong envelopes be x2
Let the no. of ways in which exactly 3 letters go into the right envelopes & 1 letter going into wrong envelope IS NOT POSSIBLE
Let the no. of ways in which all 4 letters go into right envelopes be x4 (only 1 way though)

So, the no. of ways in which all 4 letters go into wrong envelopes = 4! - (x1 + x2 + x4)

Calculating x1 itself is another problem by its own!
i.e, x1 = 4c1 * <No. of ways in which rest of the 3 letters go into 3 wrong envelopes>

And for 3 letters to go into 3 wrong envelopes, we again have 3! - (1 right letter but 2 wrong letters + all 3 right letters) i.e, 3! - (3+1) i.e., 2 ways

So, x1 = 4c1*2 = 8

Coming to x2; 2 letters going into right envelopes can be chosen by 4c2 ways and the rest 2 going wrong is only 1 possibility (interchanging other two)

So, x2 = 4c2*1 = 6

And we know x3 = 1

Hence, the no. of ways in which all 4 letters go into wrong envelopes = 4! - (8+6+1) = 4!-15 = 9 ways

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

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New post 26 Aug 2008, 04:03
ssandeepan 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


I get only 9:

Envelope A can get b,c or d:
If A gets b, B can get a,c, or d. If B gets a, then c goes in D and d in C. If B gets c, then d goes in C and a goes in D
If A gets c, B can get a or d. If B gets a, then C gets d and D gets b. If B gets d, then C and D get a and b or vice versa.
If A gets d, B can get a or c. If B gets a, then C gets b and D gets C. If B gets c, then C and D get a and b or vice versa.

Perhaps it is better to count the converse: 1 way for 4 correct, 0 ways for 3 correct, 4C2 = 6 ways for 2 correct and 8 ways for only one correct ( 4C1 multiplied by 2). 24 - 1 - 6 - 8 = 9

I like this:
Think of two sets: one with a and b, the other with c and d

3 cases:

case 1: no exchange of letters between the sets- the two sets must each do the only inward swap possible (1)
case 2: one exchange of letters between the sets 2C1 x 2C1 (choose one letter for each set for the exchange, after the exchange, two sets must each do the inward swap (4)
case 3: each letter in the first set moves to the second and vice versa- the new sets can be permuted in 2 ways each (4)
4 + 4 + 1 = 9

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

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New post 26 Aug 2008, 11:18
E1 E2 E3 E4- envelopes
LI L2 L3 L4 - letter

total possible ways = 4! = 24

preffered ways
(eg, when we have L1 in E1)
E1 E2 E3 E4
L1 L4 L2 L3
L1 L2 L4 L3
L1 L3 L2 L4
SO for 1 position of L1, we can have correspoding 3 ways which we dont want the letter arrangement. since there are 4 letter,
4 *3= 12 ways which we dont want.
so no of ways which we want = total poss ways - total unwanted ways = 24-12 = 12

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Re: PS A secretary types 4 letters and then addresses   [#permalink] 26 Aug 2008, 11:18
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