#### SEATING ARRANGEMENT

- In order to solve seating arrangement questions, first of all diagram should be made. By doing so questions are easily and quickly solved.

*Example 1:*

- 6 Boys are sitting in a circle and facing towards the centre of the circle.
- Rajeev is sitting to the right of mohan but he is not just at the left of Vijay.
- Suresh is between Babu and Vijay.
- Ajay is sitting to the left of Vijay.

Who is sitting to the left of Mohan ?

*Solution :*

Hence, Babu is sitting to the left of Mohan.

*Example 2:*

- Eleven students A, B, C, D, E, F, G, H, I, J and K are sitting in first line facing to the teacher.
- D who is just to the left of F, is to the right of C at second place.
- A is second to the right of E who is at one end.
- J is the nearest neighbour of A and B and is to the left of G at third place.
- H is next to D to the right and is at the third place to the right of I.

Who is just in the middle ?

*Solution :*

Hence, I is just in the middle.

*Example 3:*

Siva, Sathish, Amar and Praveen are playing cards. Amar isto the right of Sathish, who is to the right of Siva.

Who is to the right of Amar ?

*Solution :*

Hence Praveen is to the right of Amar.

*Example 4:*

- A, B and C are three boys while R, S and T are three girls. They are sitting such that the boys are facing the girls.
- A and R are diagonally opposite to each other.
- C is not sitting at any of the ends.
- T is left to R but opposite to C.

(A). Who is sitting opposite to B ?

(B). Who is sitting diagonally opposite to B ?

*Solution :*

(A). Hence, R is sitting opposite to B.

(B). Hence, S is sitting diagonally opposite to B.

- N! = N(N-1)(N-2)(N-3)……1
- 0! = 1! = 1
- nCr = n!/( n−r) !( r!)
- nPr = n! /(n−r) !

▪ Arrangement : n items can be arranged in n! ways

▪ Permutation : A way of selecting and arranging r objects out of a set of n objects, nPr = n!/( n−r) !

▪ Combination : A way of selecting r objects out of n (arrangement does not matter) nCr = n! /(n−r )!(r!)

▪ Selecting r objects out of n is same as selecting (n-r) objects out of n, nCr = nCn-r

▪ Total selections that can be made from ‘n’ distinct items is given ∑_{ k=0}^{n} nCk = 2^{ n}

#### Partitioning :

▪ Number of ways to partition n identical things in r distinct slots is given by ^{(n+r−1)}C_{(r−1)}

▪ Number of ways to partition n identical things in r distinct slots so that each slot gets at least 1 is given by ^{(n−1)}C_{(r−1)}

▪ Number of ways to partition n distinct things in r distinct slots is given by r^{ n}

▪ Number of ways to partition n distinct things in r distinct slots where arrangement matters = (n+r−1)! /(r−1)!

#### Arrangement with repetitions :

- If x items out of n items are repeated, then the number of ways of arranging these n items is n!/ x! ways.
- If a items, b items and c items are repeated within n items, they can be arranged in n! /(a! b! c!) ways.

#### Rank of a word :

▪ To get the rank of a word in the alphabetical list of all permutations of the word, start with alphabetically arranging the n letters. If there are x letters higher than the first letter of the word, then there are at least x*(n-1)! Words above our word.

▪ After removing the first affixed letter from the set if there are y letters above the second letter then there are y*(n-2)! words more before your word and so on. So rank of word = x*(n-1)! + y*(n-2)! + .. +1

#### Integral Solutions :

▪ Number of positive integral solutions to x1+x2+x3+…..+xn= s where s ≥ 0 is ^{(s−1)}C_{(n−1)}

▪ Number of non-negative integral solutions to x1+x2+x3+…..+xn = s where s ≥ 0 is ^{(n+s−1)}C_{(n−1)}

#### Circular arrangement :

Number of ways of arranging n items around a circle are 1 for n = 1,2 and (n-1)! for n≥3. If its a necklace or bracelet that can be flipped over, the possibilities are (n-1)! /2

###### Derangements :

If n distinct items are arranged, the number of ways they can be arranged so that they do not occupy their intended spot is D = n!(1 /0! − 1/ 1! + 1 /2! − 1 /3! +…… ( −1 )^{ n} /n! )