Dearrangement Formula | Permutations of Alike Objects

Here you will learn dearrangement formula and permutation of alike objects with example.

Let’s begin –

Dearrangement Formula

There are n letters and n corresponding envelopes. The number of ways in which letters can be placed in the envelopes (one letter in each envelope) so that no letter is placed in correct envelope is

n![1 – \(1\over 1!\) + \(1\over 2!\) +……..+\({(-1)^n}\over n!\)]

Example : A person writes letters to six friends and addresses the corresponding envelopes. In how many ways can the letters be placed in the envelope so that all the letters are in the wrong envelopes.

Solution : The number of ways in which all the letters can be placed in wrong envelopes.

= 6!(1 – \(1\over 2\) – \(1\over 6\) + \(1\over 24\) – \(1\over 120\) + \(1\over 720\))

= 720(\(1\over 2\) – \(1\over 6\) + \(1\over 24\) – \(1\over 120\) + \(1\over 720\))

= 360 – 120 + 30 – 6 + 1 = 265.

Permutations of alike objects

Case-1 : Taken all at a time

The number of permutations of n things taken all at a time: when p of them are similar of one type, q of them are similar of second type, r of them are similar of third type and the remaining n – (p + q + r) are all different is :

\(n!\over {p! q! r!}\).

Example : In how many ways the letter of the word “ARRANGE” can be arranged without altering the relative position of vowels & consonants.

Solution : The consonants in their position can be arranged in \(4!\over 2!\) = 12 ways.

The vowels in their position can be arranged in \(3!\over 2!\) = 3 ways.

\(\therefore\)    total number of arrangements = \(12\times 3\) = 36.

Case-2 : Taken some at a time

Example : Find the total number of 4 letter words formed using four letters from the word “PARALLELOPIPED”.

Solution : Given letters are PPP, LLL, AA, EE, R, O, I, D.

Case 1: All distinct,    No. of words = \(^{8}C_4\).4! = 1680

Case 2: 2 alike, 2 distinct,    No. of words = \(^{4}C_1\).\(^{7}C_2\).\(4!\over 2!\) = 1008

Case 3: 2 alike, 2 other alike,    No. of words = \(^{4}C_2\).\(4!\over 2! 2!\) = 36

Case 4: 3 alike, 1 distinct,    No. of words = \(^{2}C_1\).\(^{7}C_1\).\(4!\over 3!\) = 56

\(\therefore\)    Total no. of words = 2780

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