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Permutation and combination

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• 1. Permutation and Combination for everybody
• 2. In the waiting room of a doctor’s chamber there are three chairs. Five patients, A,B,C,D and E have to be seated on these chairs. In how many ways can we fill up three seats ?
• 3.
• The first chair can be filled in 5 ways. For every way the first chair is filled up, the second can be filled in 4 ways and so on. The total no. of ways is therefore 5 x 4 x 3 = 60.
• We can write this as
• 5x4x3x2x1 / 2x1 = 5! / 2!. This is denoted by 5 P 3 i.e. no. of permutations of 5 patients, taken 3 at a time.
• 4. Had there been n patients and r chairs, the number of ways would have been n! / (n-r)! = n P r.
• If there are 5 patients and five chairs the no. of ways will be
• 5 P 5 = 5 x 4 x 3 x 2 x 1= 5! = 120.
• In general n P n = n!
• 5. Suppose we have 5 students and you have to select three to represent your college in a debate. This can be done in 5 C 3 ways.
• If now 3 chairs are provided for the team, in how many ways can you seat the 3 students ? We know that this number is 3 P 3 = 3!.
• 6.
• There fore 5 C 3 x 3! is the number of ways in which you can fill three seats with 5 students. But that is nothing but 5 P 3.
• 5 P 3 = 5 C 3 x 3!
• 5 C 3 = 5 P 3 / 3!
• In general, n C r = n P r / r!
• = n!/(n-r)! r!
• 7. Example:
• With 8 batsmen, 2 wicketkeepers, 4 spinners and 5 pacers, in how many ways can you select a team with 5 batsmen, one wicketkeeper, 2 spinners and 3 pacers ?
• The number is 8 C 5 x 2 C 1 x 4 C 2 x 5 C 3
• = [8!/5!.3!] [2!/1!.1!] [ 4!/2!.2! ] [5!/3!.2!]
• = 56.2.6.10
• = 6720
• 8.
• Ex: 6 persons, Amal, Bimal, Chand, Diana, Emily and Farah are to be seated at a round dinner table. Amal and Diana are to sit next to each other. In how many ways can you allot seats for them ?
• We first consider Amal and Diana as one person. Five persons can be seated in 5! ways. Since the positions of Amal and Diana can be interchanged, we multiply this number by two and get 2 x 5!. However, the whole group can be rotated as: AD-B-C-E-F  F-AD-B-C-E  E-F-AD-B-C etc. which do not give new arrangements. So we divide the number by 5 and get 2 x 5!  5 = 48
• 9.
• Ex: 52 cards are distributed among 4 players. In how many ways can a player get all the aces ?
• We consider 4 aces as one card. So there are 48 other cards. A person can get 9 cards out of 48 in 48 C 9 ways. This is a very large number !