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MATHS PROJECT

MATHS PROJECT. Submitted by Neethi Raveendran 11- E K V Pattom. Permutations. &. Combinations. Fundamental Principle of Counting. Illustration:.

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MATHS PROJECT

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  1. MATHS PROJECT Submitted by Neethi Raveendran 11- E K V Pattom

  2. Permutations & Combinations

  3. Fundamental Principle of Counting Illustration: A school bag can be chosen in 2 different ways. After a school bag is chosen, a tiffin box can be chosen in 3 different ways. Hence, there are 2*3 = 6 pairs of school bag and a tiffin box. For each of these pairs a water bottle can be chosen in 2 different ways. Hence there are 6*2 = 12 different ways in which, Sabnam can carry these items to school. If we name the 2 school bags as B1,B2, the three tiffin boxes as T1,T2,T3 and the two water bottles as W1,W2, these possibilities can be illustrated in the below given figure.

  4. B1T1W1 W1 B1T1W2 T1 B1T2W1 B1T2W2 B1 B1T3W1 B1T3W2 B2T1W1 B2 B2T1W2 B2T2W1 B2T2W2 T2 W2 B2T3W1 B2T3W2

  5. Illustration : There are 12 buses running between two places. Find the number of ways in which a person can go from one place can go from one place to another and come back by a different way. A person can make onward journey in any of the 12 buses. Since, he is not coming back through the same bus, he made his onward journey, there left 11 buses for the return journey. Hence by the fundamental principle of counting the total number of ways = 12*11 There are 5 chairs in hotel. Find the number of ways in which they can take their seats. The first boy can take his seat in anyone of the 5 chairs hence in 5 days. They left 4 chairs for the second boy. They left 4 chairs and he can be seated in 4 ways. Hence the third boy can be seated in 3 ways. Hence by principle of counting three boys can take their seats in 5*4*3 = 60.

  6. If one operation can be performed in m ways (when it has been performed in anyone of these m ways) and the second operation in anyone of n ways. Then total number of ways of performing both the operations = m*n. This rule can be extended to any number of cases. Permutations : The word permutation means arrangement. The different arrangements that can be made with a given number of things taking sum of all of then at a time is called permutation. Permutations of n different objects taken r at a time is the number of ways of filling up r places with n things.

  7. PROVE THAT nPr = n!/(n-r)! SOLUTION: Number of permutations of n things taken r at a time is the same as the number of ways of filling r places with n things. The first place can be filled with any one of the n things hence in n ways. There left (n-1) things for second thing. The second place can be filled with any one of the (n-1) things. Hence the first and second can be filled with anyone of the (n-2)things.Hence in (n-2) things. By the principle of counting first three places can be filled in n(n-1)(n-2) ways. …………………………….. r 1 2 3 n n-1 n-2 n-r+1

  8. nPr = n(n-1)(n-2)(n-3)……………………………………(n-r+1) nPr = n(n-1)(n-2)(n-3)………………(n-r+1)(n-r)(n-r-1)(n-r-2).3.2.1 (n-r) (n-r-1) (n-r-2).3.2.1 nPr = n! (n-r)! Examples 1) 8P2 = 8!/6! = 56 2) 4P3 = 4!/1! = 4 3) nPn = n!/(n-n)! = n! 0! = 1

  9. nPr = n-r+1 • Proof: • = n!/(n-r)! • n!/(n-r+1)! • = n!/(n-r)! . (n-r+1)!/n! • = (n-r+1) (n-r)!/(n-r)! • = n-r+1 • nPr is defined • n, r are positive integers • n is always greater than or equal to r • r is always greater than or equal to 0

  10. PRACTICAL PROBLEMS BASED ON PERMUTATION 1)How many 5 digit distinct telephone numbers can be formed with digits 0,1,2,3,4,5,6,7,8,9. Required number of telephone numbers = No. of permutations taken 5 at a time = 10P5 = 10!/(10-5)! = 10.9.8.7.6.5!/5! = 30240 2)How many ways can the letters of the word DELHI be arranged? The word DELHI consist of 5 different ,then they can be arranged among themselves in 5P5 ways = 5! = 5.4.3.2.1 = 120

  11. Combinations The word combination means selection. The different ways in which a selection of r things can be made from n things without regard to the order of selection or arrangement are called the combination of n things r at a time. The number of combinations of n different things taken r at a time is denoted by nCr. nC1 = The no. of combinations of n diff. things taken one at a time=n nCn = The no. of combinations of n diff. things taken all at a time = 1 nCo = The no. of combinations of taken none at a time= 1

  12. nCr = n!/r!(n-1)! Prove nCr = nCn-r LHS = nCr = n! r!(n-r)! RHS = nCn-r = n!/(n-r)!(n-n+r)! = n!/r!(n-r)! nCr + nCr-1 = n+1 Cr

  13. nCr/nCr-1 = n-r+1/r Practical problems based on combination • How many different teams of 7 players can be chose from a group of 10 players? • Total no of combinations = 10!/7!.3! • = 10.9.8.7!/7!.3! • = 120 • 2) There are 5 red balls, 7 white balls and 3 black balls. Find the number of ways in which 3 balls can be selected so that each of them has different balls? • Since the 3 balls are of different colours, we have to select one ball from each colour. One red ball from 5 can be selected in 5C1 ways. One white ball from 7 can be selected in 7C1 ways. One black ball from 3 black balls can be selected in 3C1 ways.

  14. Total number of ways = 5C1 + 3C1 + 7C1 = 5+3+7 =15 3) A committee of 5 persons is to be selected from among 6 boys and 5 boys. Find the number of ways in which committee can be selected so that boys are majority? Since the committee consists of boys majority, the selection can be done in the following manner Boys Girls 5 0 4 1 3 2

  15. Total number of ways = 6C5 . 5C0 + 6C4.5C1 + 6C3.5C2 = 6C1 + 6C2.5 + 6C3.5C2 = 6+75+200 =281 4) The letters of the word FATHER be permitted and arranged as in a dictionary (with or without meaning). Show that its rank is 261. Arrange them in the alphabetic order like A,E,F,H,R,T. No. of words starting with A= 1. 5P5 No. of words starting with E= 1.5P5 No. of words starting with FAE=1.1.1.3P3 No. of words starting with FAH=1.1.1.3P3 No. of words starting with FAR=1.1.1.3P3 No. of words starting with FATE= 1.1.1.1.2P2

  16. Adding all these we will get, =5!.2 + 3!.3 + 2! =120*2 + 6*3 + 2 = 240+18+2 = 260 Therefore FATHER is the 261 st word.

  17. CONCLUSION The concepts of permutations and combinations can be tracked back to the advent of Jainism in India. The credit however goes to Jains. Outside India these developments started with humble beginnings. The use of permutation and combination thus became an integral part in our daily life. Here, we dealt with the techniques of determining ways of arranging different number of objects and selecting different types of objects. As it is clear now that it is the basic fundamental to learn these all.

  18. THANK YOU !!!!!

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