Week 9 – SS2 Second Term Further Mathematics Notes

 WEEK NINE
TOPIC:PERMUTATION AND COMBINATION :
PERMUTATION:Definitoin, Concept , Different Arrangement Of Items, Cyclic Permutation .

1. Definition, Concept:

   Definition: Permutation is defined as the number of arranged of objects. The different orders of arrangement are important. E.g. Find the number of ways of arranging the letters pqr.
   Pqr, prq, qrp, qpr, rqp. The number of ways is 6 ways
   Similarly, for 4 letters the number of arrangement is 24
   In general, the number of different arrangement of n different objects is equal to n! (n factorial)
   N! = n x (n-1) x (n-2) x … x 3×2 x 1×0!    (But, 0! = 1)

2. Simplify the following: A. 5!        B.

   Solution
   

3. 5! = 5x4x3x2x1 = 120
4. = 7x6x5x4! = 7×5 = 35
5. Find the number of ways of arranging the letters of the word MACHINE

   Solution:
   There are seven different letters in the word MACHINE, therefore the number of permutation is 7! = 7x6x5x4x3x2x1 = 5040 ways

6. Simplify (n + 1)! = (n+1)n! = n+1

   (n-1)!        (n-1)n!    n-1

    ARRANGEMENT OF n-OBJECTS TAKING r-OBJECTS
   If we are interested in the number of ways 2 letters of a 4 lettered word cnbe arranged, then the np_r is the permutation of n objects taking at a time
   ⁿp_r =
   Example: Evaluate:     (a) 8p₃        (b) 11p₉
   Solution_:
   

7. ⁸P₃ = = = = 8x7x6 = 336
8. ¹¹P₉= = 11x0x9x8x7x6x5x4x3x2! = 19958400
9. In how many ways can three people be seated on eight seats in a row?

   Solution:
   1^st seat can be occupied by any of the 8 = 8 ways
   2^nd seat can be occupied in 6 ways
   Hence, the number of ways = 8x7x6 = 336 ways
   Alternatively, n = 8, r = 3
   ⁿP_r = = = = 8x7x6 = 336 ways

    EVALUATION
   

10. In how may ways can 8 students be seated in a row?
11. In how many ways can the 1^st, 2^nd 3^rd prizes be won by 6 athletetes in a race?
12. In how many ways can the letters of the word HISTORY be arranged?

     CYCLIC PERMUTATION: Cyclic permutation is the arrangement of things around a circular object. Since a circular table has no beginning and no end, the number of arrangement is 1 x (n – 1) !
    If the circular object can be turned over e.g. circular ring e.t.c. the number of arrangement =
    Example: In how many ways can 6 members of a disciplinary committee be seated round a circular table?

     Solution:
    The number of ways = (n – 1)! X 1
    N = 6,
    Hence, (6 – 1)! X 1 = 5! X 1 = 120 ways
    PERMUTATION OF IDENTICAL OBJECTS:
    The number of ways of permuting n objects taking n at a time with n, objects alike, n₂alke is,

13. Find the number of ways the word MATHEMATICS can be arranged.

    Solution:

     MATHEMATICS
    There are: 2Ms, 2Asm 2Ts and 11 letters.
    N=11, n1 = 2! N2 = 2! N3 = 2
    = 11x10x9x7x6x5x4x3x2x1 = 4989600

     CONDITIONAL PERMUTATION:
    Sometimes restrictions are placed on the order of arrangements of objects
    Examples:
    

14. Find the number of ways the letters of the word COMMITTEE can be permuted, if the 2Ts must always be together.

    Solution:
    The 2Ts must be together, we can lump them as follows: COMMI (TT) EE = 8!
    = = 10080 ways

15. Find the number of ways of arranging the letters of the word MOSHOESHOE if the letter M must always begin a word

    Solution:
    Since letter m must always begin, and then m can only occupy the first position
    i.e M = 1 way
    other letters, OSHOESHOE = = 9x7x6x5x4 = 7560 ways

     COMBINATION: Selection, Conditional Selection And Its Application
    Combination can be defined as the number of ways r – objects can be selected from n – objects irrespective of the arrangement
    Hence, the notation is thus, ⁿC_r or (ⁿ_r)
    , ⁿC_r=
    Relationship between permutation and combination is thus, nCr =
    Example:
    

16. Evaluate ¹⁰C₄

    Solution:
    ¹⁰C_{4 =} = 10 x 3 x 7 = 210
    

17. In how many ways can three books be selected from 12 books?

    SOLUTION:
    N = 12, r = 3, ¹²C₃ = = 2x11x10 = 220 ways

    1. A committee consisting of 3 men and 5 women is selected from 5 men and 10 women. Find how many ways this committee can be formed.

    Solution:
    MEN                    WOMEN
    R = 3, n = 5                r = 5, n = 10
    ⁵C₃ =      = 10            ¹⁰C₅ = = 252
    Therefore number of ways of selecting the committee = 10×252 = 2520 ways.

     GENERAL/ REVISION EVALUATION
    

18. Find the number of ways the letters of the word FURTHER can be arranged
19. Find the number of ways of arranging 7 people in a straight line, if two particular people must always be separated
20. In how many ways can 6 pupils be lined up if 3 of them insist in the following one another
21. Verify that = (n – 1) (n – 2) (n – 3)!

     READING ASSIGNMENT
    Read permutation and combination, further mathematics project 2 pages 47-54

     WEEKEND ASSIGNMENT
    

    1. Evaluate ⁶C₂ + ⁶C₃ + ⁶C₄ + ⁶C₅ (a) ⁶C₆ (b) ⁶C₅ (c) ⁸C₅
    2. How much ways can the letters of the word EVALUATE be arranged? (a) 10080 (b) 20160 (c) 40320
    3. In how many ways can 2 boys and 3 girls be arranged to sit in a row, if the boys must sit together (a) 6 (b) 4 (c) 24
    4. Find the number of ways 6 people can be seated in a round table, if two particular friends must sit next to each other (a) 48 9b) 24 (c) 120
    5. In how many ways can 6 pupils be lined up if 3 of them insist on following one another? (a) 720 (b) 144 (c) 24

     THEORY
    

    1. Out of 7 lawyers, 5 judges, a committee consisting of 3 lawyers, 2 judges is to be formed, in how many ways can this be done, if
       1. Any lawyer and any judge can be included
       2. One particular judge can be included
       3. Two particular lawyer cannot be in committee
22. If ⁿP₃ / ⁿC₂ = 6, find the value of n