counting problem

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counting problem

by sairakarim07 » Tue May 07, 2013 11:29 am
I want to know Dearangement of permutation to solve following problem.

Tanya prepared four different letters to be sent to four different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the envelopes at random, what is the probability that only one letter will be put into the envelope with its correct address?

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by srcc25anu » Tue May 07, 2013 11:48 am
Total # of arrangements possible to put 4 letters in 4 envelopes = 4! = 24

If exactly one letter is put in correct envelope, those cases are:

A C D B
A D B C
D B A C
C B D A
B D C A
D A C B
B C A D
C A B D

Total 8 such cases when only 1 letter is put in the correct envelope.

Reqd prob = 8/24 = 1/3

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by srcc25anu » Tue May 07, 2013 11:55 am
Just some further analysis of this problem:

Had it been to find the # of arrangements with exactly 2 letters put in the correct envelope, that would equal 4C2 = 6 ways

A B D C
A C B D
A D C B
D B C A
C B A D
B A C D

# ways where all letters in their correct envelopes: 1 (A B C D)

# of ways where no letter in correct envelope = 24 - (All correct) - (1 correct) - (2 correct)
= 24 - 1 - 8 - 6 = 9 ways

Another way to find # ways where NO letter is put in its correct envelope = 4! * [1/2! - 1/3! + 1/4!] = 9

General formula: n! * [1/2! - 1/3! + 1/4! - .... 1/n!]