De Arrangement

8/22/2019 De Arrangement

http://slidepdf.com/reader/full/de-arrangement 1/3

Funda 1: De-arrangement

If ‘n’ distinct items are arranged in a row, then the number of ways they can be rearranged suchthat none of them occupies its original position is,

n! * ((1/0!) – (1/1!) + (1/2!) – (1/3!) + … ((-1)

n

/n!))

Note: De-arrangement of 1 object is not possible.

Dearr(2) = 1; Dearr(3) = 2; Dearr(4) =12 – 4 + 1 = 9; Dearr(5) = 60 – 20 + 5 – 1 = 44

Example,

A person has eight letters and eight addressed envelopes corresponding to those letters. In how

many ways can he put the letters in the envelopes such that exactly 5 of them get delivered

correctly?

Solution,

At first, select the five letters that get delivered correctly. That can be done in8C5 ways.

Now, the other three must get delivered to the wrong address. That can be done in Dearr(3) = 2

ways.

So, total ways is 2 x8C5 = 2 x 56 = 112 ways.

Funda 2: Partitioning

‘n’ identical items

in ‘r’ distinct

groups

No restrictions:n+r-

Cr-1

No group empty:n-1

Cr-1

‘n’ distinct objectsin ‘r’ distinct

groups

No restrictions: r n

Arrangement in a group is important: (n + r -1)! / (r-1)!

Note: Other than standard distribution / partitioning problems, these ideas can be used to solve

questions in which number of solutions are asked.

Example,

How many solutions are there to the equation a + b + c = 100; given that

a) a, b and c are whole numbers.

b) a, b and c are natural numbers.



Solution,

Case a) is identical to a case in which 100 identical chocolates are being distributed in three kidsa, b and c. It is possible that one kid gets all the chocolates. In this case, we will use the formula

for distributing ‘n’ identical items in ‘r’ distinct groups where n = 100 and r = 3.

So, it can be done in102

C2 ways.

Case b) is identical to one in which 100 identical chocolates are being distributed in three kids a,

b and c. Every kid must get at least one chocolate. In this case, we will use the formula for

distributing ‘n’ identical items in ‘r’ distinct groups where no group is empty and n = 100 and r =3.

So, it can be done in99

C2 ways.

Example,

In how many ways can you distribute 5 rings in

a) 4 boxes

b) 4 fingers

Solution,

First of all we need to identify the difference between distributing in boxes and distributing in 4fingers. The distinction is that in case of fingers, unlike boxes, the order in which rings are

placed matters.

In Case a; Ring 1 can go in any of the four boxes, so it has four choices. Ring 2 can also go in

any of the four boxes, so it has four choices. Similarly for Ring 3, Ring 4 and Ring 5; there are 4

choices each. So, the total number of ways of distribution is = 4 x 4 x 4 x 4 x 4 = 45. This is

essentially how the formula r n

is derived.

In Case b) Ring 1 can go in any of the four fingers, so it has 4 choices. Ring 2 can go in any of the four fingers but it has five choices. There is a finger, say F3, which contains the ring R1.

Now, on F3, R2 has two choices – it can go above R1 or below R1. So, the total number of

choices for R2 is 5.

Ring 3 can go in any of the four fingers but it now has 6 choices.

Ring 4 can go in any of the four fingers but it will now have 7 choices.Ring 5 can go in any of the four fingers but it will now have 8 choices.

So, the total number of way of distribution of rings is = 4 x 5 x 6 x 7 x 8 = 8! / 3!

This is essentially how the formula (n + r -1)! / (r-1)! is derived.

Funda 3



Number of ways of arranging ‘n’ items, out of which ‘p’ are alike, ‘q’ are alike and ‘r’ are alike

given that p + q + r = n

OR

Number of ways of distributing ‘n’ distinct items, in groups of size ‘p’, ‘q’ and ‘r’ given that p +q + r = n

Is equal to,

n! / (p! * q! * r!)

I hope that this would help you solve problems in the exam. May be the chocolate you end up

getting is a Bournville. Maybe you would have earned it.

Documents

De Arrangement